Data sufficiency questions have five standard options. They are listed below and will not be repeated for each question.
(A) Statement (1) ALONE is sufficient, but statement (2) ALONE is not sufficient to answer the question asked.
(B) Statement (2) ALONE is sufficient, but statement (1) ALONE is not sufficient to answer the question asked.
(C) both the statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked.
(D) EACH statement ALONE is sufficient to answer the question asked.
(E) Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed.
101. A man has three sons, among whom he distributed his wealth. Is the share of the eldest son greater than $12000?
(1) The eldest son received the greatest share.
(2) The total wealth was $30000.
Solve yourself:
102. Isx > y?
(1) 2x + 3y − z = 0 (2) 7x − 3y − 2z = 0
Solve yourself:
103. Ifz is positive, is x > z?
(1) 2x + 3y − z = 0 (2) 7x − 3y − 2z = 0
Solve yourself:
104. What is Harry’s present age?
(1) When Harry joined school with Ron, five years ago, sum of their ages was 13 years.
(2) When Harry joined school with Ron, five years ago, Harry was one year younger to Ron.
Solve yourself:
105. What is Harry’s present age?
(1) When Harry joined school, five years ago, he was half as old as his elder brother.
(2) The age of Harry’s elder brother is five years less than twice Harry’s age.
Solve yourself:
106. A man purchased few oranges and few apples. What is the total price of one orange and one apple?
(1) The man purchased 30 fruits, consisting of only apples and oranges and the total price came to $40.
(2) Had the man purchased as many apples as oranges and as many oranges as apples, he would have paid $80.
Solve yourself:
107. What is the sum of digits of a four-digit numberabcd, where a is the first digit, b is the second digit,c is the third digit and d is the fourth digit and the sum of the first three digits equals the fourth digit?
(1) The sum of its third digit and twice its second digit equals 10 times its first digit.
(2) The sum of the first and last digits equals 5 times the second digit.
Solve yourself:
108. The individual ages of Ann and Bob are whole numbers such that each of them is a two-digit number. What is Ann’s age?
(1) Ann’s age is such that if the digits of her age are reversed, Bob’s age is obtained.
(2) Bob’s age exceeds Ann’s age by 1
11 of the combined ages of Ann and Bob.
Solve yourself:
109. The price of an apple is $1 and the price of an orange is $2. A man purchased few apples and few oranges. What is the number of oranges and apples purchased by the man?
(1) The man spent $41 to purchase the fruits.
(2) Had the man purchased as many apples as oranges and as many oranges as apples, he would have saved half the cost of an orange.
Solve yourself: (2) p and q are positive integers.
Solve yourself:
112. What is the largest possible integer value ofk?
(1) The roots ofx2−kx + 35 = 0 and the roots of x2−7x + k = 0 are real numbers.
(2) k > 0
Solve yourself:
113. Is 1
x + 1+ 1
y + 1 = 2a a + b? (1) ax2=by
(2) ay2=bx
Solve yourself:
114. P, Q, R and S are four cakes, is the total cost of P and S less than that of Q and R?
(1) The cost of R is half of the cost of S.
(2) The cost of Q is twice that of R, which in turn, is costlier than P.
Solve yourself:
115. Ifm and n are integers, what is the largest possible value of m?
(1) 1 m + 2
n = 1 10
(2) m and n are positive integers.
Solve yourself:
116. A, B and C have some marbles with them. Does C have at least 50 percent of all the marbles with them together?
(1) The sum of the number of marbles with A and B together is 40 percent less than that with B and C together.
(2) The sum of the number of marbles with B and C together is 20 percent less than that with A and C together.
Solve yourself:
117. A group of friends wanted to purchase a camera by sharing the cost equally. What is the price of the camera?
(1) The price of the camera is between $210 and $230.
(2) If two friends back out, the remaining friends have to contribute $1 extra to purchase the camera.
Solve yourself:
118. A man has a certain amount of money in $1 and $10 bills. Can the total number of bills with him be nine?
(1) The number of $1 bills multiplied by the number of $10 bills is equal to the total money (in dollars) with him.
(2) The number of $1 bills is greater than eight.
Solve yourself:
119. The letters A, B, C, D and E represent distinct numbers among 2, 4, 5, 6 and 10, not necessarily in the same order. What is the value of B?
(1) A + C = E (2) A + E = B
Solve yourself:
120. Ifx, y and z are positive numbers and 5x + 7y + 8z = k, can the value of k equal 20?
(1) 2x + 3y + 4z = 10 (2) 3x + 4y + 5z = 12
Solve yourself:
121. Can the value ofk equal 20?
(1) 5x + 7y + 9z = k
(2) x, y and z are positive integers
Solve yourself:
122. Iff (x) = x3−4x + p, is 0 < p < 3?
(1) f (0) × f (1) < 0 (2) f (0) > 0
Solve yourself:
123. Iff (x) = (a − xn)
1
m
andf (1) = p, what is the value of f p?
(1) n = m (2) a = 1
Solve yourself:
124. a, b, x, y, . . .
Each term of the above sequence is 9 more than 1
3 of the previous term. What is the value of x − y?
(1) a = 54 (2) y = 5
6x Solve yourself:
125. Iff(n+1)(x) = f(n)(x) + 1, for all positive integer values of n = 1, 2, 3, . . . , what is the value of f(4)(x)?
(1) f1(x) = 0 (2) f5(x) = 4
Solve yourself:
126. If f (n + 3) = f (n + 2) × f (n), for all positive integer values of n = 1, 2, 3, . . . , what is the value off (5) − f (3)?
(1) f (1) = f (2) = 1 (2) f (3) = −1
Solve yourself:
127. Aluminum costs $2 per kilogram, and copper costs $4 per kilogram. If 10 kilograms of alloy K consists ofx kilograms of aluminum and y kilograms of copper, is x > y?
(1) y > 3
(2) The cost of 10 kilograms of alloy K is less than $30.
Solve yourself:
128. What is the age of P, if the average (arithmetic mean) age of P, Q, and R is 24 years?
(1) The difference between the ages of P and Q is 6 years and R is 6 years elder to Q.
(2) Among the three, P is the youngest and R is the oldest.
Solve yourself:
129. What is the age of P, if the average (arithmetic mean) age of P, Q, and R is 24 years and no two among the three have the same age?
(1) The difference between the ages of P and Q is 6 years and R is 6 years elder to Q.
(2) Among the three, P is the youngest and R is the oldest.
Solve yourself:
130. A, B and C have a total of $60 with them. Does A have the highest amount?
(1) A and B together have 40 percent more than what C has.
(2) B has $6 more than what A and C together have.
Solve yourself:
131. Is the present age of B more than double the age of A?
(1) B’s age was double the age of A, four years back.
(2) B is 20 years older than A.
Solve yourself:
132. Jill bought only apples worth $0.30 each and oranges worth $0.58 each. How many apples did she buy?
(1) She bought $8.80 worth of apples and oranges.
(2) She bought an equal number of apples and oranges.
Solve yourself:
133. Juan bought some paperback books that cost $8 each and some hardcover books that cost $25 each. If Juan bought more than 10 paperback books, how many hardcover books did he buy?
(1) The total cost of the hardcover books that Juan bought was at least $150.
(2) The total cost of all the books that Juan bought was less than $260.
Solve yourself:
134. Some computers at a certain company are Brand X and the rest are Brand Y. If there are a total of 880 computers at the company, how many of the computers are Brand Y?
(1) Ratio of the number of Brand Y computers to the number of Brand X computers at the company is 5 to 6.
(2) The number of Brand X computers is greater than the number of Brand Y computers at the company by 80.
Solve yourself:
135.
+ x y z
d p m
e q n
f
The figure above represents an addition table where four entries,p, q, m and n are shown; for example,d + x = p. What is the value of (m + n)?
(1) d + y = −3 (2) e + z = 12
Solve yourself:
136. The total charge to rent a car for one day from Company J consists of a fixed charge of $15.00 and a charge of $0.20 per mile driven. The total charge to rent a car for one day from Company K consists of a fixed charge of $20.00 and a charge of $0.10 per mile driven. Is the total charge to rent a car from Company J for one day and drive it forx miles less than $25.00?
(1) The total charge to rent a car from Company K for one day and drive it forx miles is less than $25.00
(2) x < 50
Solve yourself:
137. What was the cost of a certain telephone call?
(1) The call lasted 8 minutes.
(2) The cost of the first minute of the call was $0.32, which was twice the cost of each minute of the call after the first.
Solve yourself:
138. Yesterday, Nan parked her car at a certain parking garage that charges more for the first hour than for each additional hour. If Nan’s total parking charge at the garage yesterday was $3.75, for how many hours of parking was she charged?
(1) Parking charges at the garage are $0.75 for the first hour and $0.50 for each additional hour or fraction of an hour.
(2) If the charge for the first hour had been $1.00, Nan’s total parking charge would have been
$4.00.
Solve yourself:
139. A piece of chalk, 8 cm long is broken into three pieces whose lengths, in cm, are distinct integers.
What is the product of the length of the three pieces?
(1) The length of the longest piece is equal to the sum of the lengths of the other two pieces.
(2) The length of the shortest piece is 1 cm.
Solve yourself:
140. Barbara sells two products, A and B. She earns commissions of $12 per unit on product A sold and $5 per unit on product B sold. If her total commission was $300, how many units of product A did she sell?
(1) Her commission from the sale of product A was at least $120.
(2) She sold 8 more units of product A than units of product B.
Solve yourself:
141. What is the total cost of 1 apple, 1 orange and 1 lemon?
(1) The total cost of 5 apples, 4 oranges and 3 lemons is Rs. 130 (2) The total cost of 3 apples, 4 oranges and 5 lemon is Rs. 110
Solve yourself:
142. If the sum of the ages of A and B is 70 years, what is the age of A?
(1) A, at present, is twice as old as B wasx years ago.
(2) B, at present, has the same age as A hadx years ago.
Solve yourself:
143. Ifx and y are positive integers, what is the value of xy?
(1) 2x+2y =8 (2) 23x+2y =32
Solve yourself:
144. Joe purchased a number of cakes priced at $13 each and a number of biscuit packets priced at
$7 each. What is the total number of cakes and biscuit packets purchased by Joe?
(1) The total price of items purchased is $33.
(2) If the count of cakes and the count of biscuit packets purchased had been interchanged, the total price of the items would have been $27.
Solve yourself:
145. Three friends, A, B and C decided to have a beer party. If each of the three friends consumed equal quantities of beer, and paid equally for it, what was the price of one beer bottle?
(1) A, B and C brought along 4, 6 and 2 bottles of beer, respectively; all bottles of beer being identical.
(2) C paid a total of $16 to A and B for his share.
Solve yourself:
146. In an experiment withn bacteria, it was found that each bacteria weighed 10−12grams. Each of then bacteria gave birth to n new bacteria, each of which also weighed 10−12grams. What was the value ofn?
(1) The firstn bacteria weighed 1
16 of the total weight of all bacteria.
(2) The total weight of all bacteria was 24 10−11
grams.
Solve yourself:
147. Ifp is an integer, what is the value of p?
(1) pq = 4 (2) q − 2p = 7
Solve yourself:
148. Ifa2−b2=20, what is value ofa?
(1) a and b are positive integers.
(2) a + b = 10
Solve yourself:
149. What is the value of a b? (1) a2
b +b = 2a
(2) (a − 2)2+ |b − 2| = 0
Solve yourself:
150. What is the value of
1 a+ 1
b
?
(1) 1 −a
a + 1 −b b =2 (2) ab
a + b = 1 4 Solve yourself:
Answer-key
4.1 Problem Solving
(68) E (69) C (70) D (71) A (72) D (73) C (74) D (75) B (76) D (77) D (78) C
(79) A (80) C (81) D (82) B (83) E (84) A (85) C (86) C (87) B (88) D (89) D
(90) B (91) A (92) E (93) B (94) D (95) C (96) A (97) A (98) B (99) D (100) C
4.2 Data Sufficiency
Solutions
5.1 Problem Solving
1. Let the price of 1 chocolate and 1 pen be $x and $y, respectively.
Thus, we have:
19x + 21y = 29 . . . (i) 21x + 19y = 31 . . . (ii) Adding (i) and (ii):
40x + 40y = 60
=> x + y =3 2 . . . (iii) Subtracting (i) from (ii):
2x − 2y = 2
=> x − y = 1 . . . (iv) Adding (iii) and (iv):
2x = 1 +3 2 = 5
2
=> x = 5 4=1.25
Thus, the price of a chocolate is $1.25 The correct answer is option C.
2. We know that the average age of the Abe and his two children ten years ago was 20 years.
Thus, at present, the age of each must have increased by 10 years; hence their average age must have been 10 + 20 = 30 years.
Thus, the present total age of Abe and his two children = 30 × 3 = 90 years.
Since Abe’s age is equal to the sum of the ages of his two children, we have:
Abe’s age + Sum of ages of his two children = 90 years
=> 2 × (Abe’s age) = 90 years
=> Abe’s age = 90
2 = 45 years.
Alternate approach:
Let the present age of Abe’s son =x years.
Present age of Abe’s daughter = 12 years.
Thus, Abe’s present age = (x + 12) years.
Thus, total present age of Abe and his two children = (2x + 24) years.
Ten years back, each must have been 10 years younger, thus the total age would have been 10
×3 = 30 years less.
Thus, total age of Abe and his two children = (2x + 24 − 30) = (2x − 6) years.
Since their average age was 20 years, their total age = 20 × 3 = 60 years.
Thus, we have:
2x − 6 = 60 => x = 33
Thus, Abe’s present age = (x + 12) = 45 years.
The correct answer is option D.
3. Let the cost of one apple, one guava and one banana be $x, $y and $z, respectively.
Thus, we have:
3x + 3y + 4z = 10 . . . (i) 3x + 2y + 4z = 9 . . . (ii)
We need to determine the value of 9x + 8y + 12z.
Adding (i) and (ii):
6x + 5y + 8z = 19 . . . (iii) Adding (i) and (iii):
9x + 8y + 12z = 29 Alternate approach:
We can see that if (i) is multiplied by 3:
9x + 9y + 12z = 30
=> 9x + 8y + 12z = 30 − y, i.e. the required value is less than 30.
Similarly, if (ii) is multiplied by 3:
9x + 6y + 12z = 27
=> 9x + 8y + 12z = 27 + 2y, i.e. the required value is greater than 27.
Thus, the required answer lies between 27 and 30, and hence, must be 29 (the only possible option).
The correct answer is option B.
4. Initial amount =a dollars, b cents.
Amount spent = $3.50 = 3 dollars, 50 cents.
Final amount = 2b dollars, 64 cents.
There are two possibilities:
• b > 50: b − 50 = 64 => b = 114 However, we know thatb < 100 Hence, this scenario is not possible.
• b < 50: Since b is less than 50, in order to properly subtract 50 cents from b, we need to borrow 100 cents froma dollars, so that the initial sum of money becomes (a − 1) dollars and (100 + b) cents.
Thus, we have: 100 +b − 50 = 64 => b = 14
Considering the subtraction of the dollars:
(a − 1) − 3 = 2b
=> a − 4 = 2 × 14 = 28
=> a = 32 Thus, we have:
Initial amount = $32.14, amount spent = $3.50, final amount = $28.64 Thus, we have:
a + b = 32 + 14 = 46.
The correct answer is option D.
5. Let the fraction be x y. Thus, we have:
x + 4 y + 4 = x
y + 1 8
=> x + 4 y + 4− x
y = 1 8
=> 4 y − x
Dividing (i) by (ii):
y − 2
Thus, the required fraction = x y = 3
4. Alternate approach:
We can use the options and find the value of the fraction that satisfies the given condition.
For example, working with Option B:
Assuming the original fraction to be3 4: The correct answer is option B.
6. We solve the first two equations and substitute the values ofx and y in the third equation.
Thus, we have:
7x − 2y = 12 . . . (i) 4x + y = 9 . . . (ii) 2x + 5y = K . . . (iii)
Multiplying (ii) by 2 and adding with (i):
7x + 8x = 12 + 18 => x = 2 Substitutingx = 2 in (ii):
8 +y = 9 => y = 1
Substitutingx = 2 and y = 1 in (iii):
K = 2 × 2 + 5 × 1 = 9.
The correct answer is option A.
7. We have:
2x + 3y = 7 . . . (i) 5x + 3y = 13 . . . (ii)
x A = y
B = 1 C . . . (iii) Subtracting (i) from (ii):
3x = 6 => x = 2 Substitutingx = 2 in (i):
3y = 7 − 4 => y = 1
Substitutingx = 2 and y = 1 in (iii):
2 A = 1
B = 1 C
SinceA, B and C are positive integers and the greatest common divisor (GCD) of A, B and C is 1, we can see that the only possible sets of values are:A = 2, B = C = 1
(Note: Other values: A = 4, B = 2, C = 2; etc. (not possible since the GCD should be 1); A = 1
2, B = 1 4, C = 1
4; etc. (not possible sinceA, B, C are positive integers).
=> A + B + C = 2 + 1 + 1 = 4.
The correct answer is option D.
8. Let the two-digit number be N = 10x + y, where x and y are the tens and unit digit of N,
Note: One should NOT solve the equations 10x +y = 32 and x +y = 5 and determine the values ofx and y, since, finally, the value of 10x + y has to be determined.
Observe that: 10x + y = 32 = 10 × 3 + 2 => x = 3 and y = 2 satisfies x + y = 5.
Thus, the product of the digits of the number = 3 × 2 = 6.
Alternate approach:
Since the sum of digits of the number is 5, the possible numbers are:
50, 41, 32, 23 and 14
We know that the ratio of 20 less than the number and 12 more than the number is 3 11. Working with the above numbers:
• N = 50 => 50 − 20 50 + 12 = 30
62 6= 3
11 – Does not satisfy
• N = 41 => 41 − 20 40 + 12 = 21
52 6= 3
11 – Does not satisfy
• N = 32 => 32 − 20 32 + 12 = 12
44 = 3
11 – Satisfies Thus the product of the digits = 3 × 2 = 6 The correct answer is option B.
9. We have:
3x + 7y = 11z . . . (i)
6x − y = 7z . . . (ii) 3x + y = kz . . . (iii)
Multiplying (i) with 2 and subtracting (ii) from it:
14y − −y = 22z − 7z
=> y = z
Substitutingy = z in (ii):
6x = 7z + y = 7z + z
=> x = 4 3z
Substitutingx = 4
3z and y = z in (iii):
3
4 3z
+z = kz
=> 5z = kz
=> k = 5.
The correct answer is option E.
10. Let the number of children in the group ben.
Let the number of pens initially with Ann bex.
If Ann takes 1 pen from each of the other children, she would have taken (n − 1) pens from the others, and hence, the total number of pens with her = (x + n − 1).
Since the number of pens with her would be thrice the number of children in the group, we have:
x + n − 1 = 3n
=> x = 2n + 1 . . . (i)
Since the total number of pens withn children is 42, and Ann alone has x = (2n + 1) pens, the number of pens with the other (n − 1) children = 42 − (2n + 1) = (41 − 2n).
Since Ann has the greatest number of pens, none of the other children can have (2n + 1) pens.
Working with the options:
Statement I:n = 5
Number of pens with Ann = 2n+1 = 11 Number of pens with the other 4 children = 41−2n = 31 Assuming that each of the 3 children has a minimum of 1 pen, then the 4th child would have 31 − 3 = 28 pens, which is greater than the number of pens (11) Ann has. – Does not satisfy
Statement II:n = 9
Number of pens with Ann = 2n+1 = 19 Number of pens with the other 8 children = 41−2n = 23 If any one other child has 19 pens, then the number of pens with the remaining 7 children would be 23 − 19 = 4, which is not possible since each child should have at least one pen. Thus, Ann definitely has the greatest number of pens. – Satisfies
Statement III:n = 15
Number of pens with Ann = 2n+1 = 31 Number of pens with the other 14 children = 41−2n = 11 Since each child must have at least one pen, we cannot have a scenario where 14 children have a total of 11 pens with them. – Does not satisfy
Note: Though it appears while working with the first two options, that the value ofn should be on the higher side, it is not prudent to assume that the largest value ofn in the options would also satisfy the conditions.
The correct answer is option B.
11. The distance covered by the ball in thetthsecond
= (Distance covered int seconds) – (Distance covered in (t − 1) seconds) = 15t2−15(t − 1)2
=15n
t2−(t − 1)2o
=15 (t − (t − 1)) (t + (t − 1))
=15 (2t − 1) Thus, we have:
15 (2t − 1) = 225
=> 2t − 1 = 15
=> t = 8
The correct answer is option D.
12. Height of the ball above the ground
=3 + 24t − 4t2
= −4 t2−6t + 3
= −4 t2−2 ×t × 3 + 3
= −4 t2−2 ×t × 3 + 32−32 + 3
= −4n
(t − 3)2−9o +3
= −4(t − 3)2+39
Since (t − 3)2is a perfect square term, it is always non-negative. Thus, when it is multiplied with (−4), it becomes non-positive.
Thus, the height of the ball from the ground is 39, reduced by the amountn
−4(t − 3)2o . Thus, the height of the ball above the ground will be the maximum when the negative term, i.e.
n−4(t − 3)2o
becomes zero, which happens when: −4(t − 3)2=0 => t = 3 The corresponding value of the height (maximum value) is thus, 39 feet.
The correct answer is option E.
13. ForN units sold at price P , total selling price (in dollars) = P × N
=P (120 − P )
Total cost incurred = $2000
Since there is neither profit nor loss, we have:
Cost price = Selling price
=> 2000 = P (120 − P )
=> P2−120P + 2000 = 0
=> P2−100P − 20P + 2000 = 0
=> (P − 100) (P − 20) = 0
=> P = 20 or 100
Thus, the maximum selling price = $100.
The correct answer is option D.
14. In a quadraticax2+bx +c = 0, the roots are real and unequal, if the discriminant, i.e. b2−4ac >
0.
Thus, for the quadraticx2−px + 8p = 0, we have:
−p2
−4 (1) 8p > 0
=> p2−32p > 0
=> p2−2 ×p × 16 + 162−162> 0
=> p − 162
Thus, the minimum positive integer value ofp = 33.
The correct answer is option E.
15. We know thatx = 3
2 satisfies the equationx2−px + 12 = 0 Substitutingx = 3
2in the equation, we have:
3
The correct answer is option C.
16. To determine the point wherey = x2+kx + l intersects the Y-axis, we need to substitute x = 0.
Since the point of intersection on the Y-axis is (0, 64), we have: 0 + 0 + l = 64
=> l = 64 . . . (i)
We also know thaty = x2+kx + l intersects the X-axis at (4, 0).
Thus, substitutingx = 4 and y = 0 in the equation, we have:
42+4k + l = 0 Substitutingl = 64:
16 + 4k + 64 = 0
=> k = −20
The correct answer is option C.
17.
12 𝑥 8
𝑥
Area of the frame
= (Area of outer rectangle) – (Area of inner rectangle)
=(12 + 2x) (8 + 2x) − 12 × 8
=24x + 16x + 4x2
=40x + 4x2
Since the area is 44 square inches, we have:
40x + 4x2=44
=> x2+10x − 11 = 0
=> (x + 11) (x − 1) = 0
=> x = −11 or 1
=> x = 1 (Negative value for the width is not possible) Thus, the required perimeter
= (Perimeter of outer rectangle) + (Perimeter of inner rectangle)
=2 {(12 + 2x) + (8 + 2x)} + 2 (12 + 8)
=2 (14 + 10) + 40
=98
The correct answer is option A.
18. We know that the height of the arch above the ground is zero at the two ends of the arch, as shown in the diagram below:
30 feet
Parbolic arch
𝑑 = 0 𝑑 = 20
Thus, we have:
h = 0 => −kd (d − 20) = 0
=> d = 0 or 20
The arch reaches its maximum height of 30 feet at the middle, i.e. ford = 0 + 20 2 =10 Thus, we have:
d = 10, h = 30
=> h = −kd (d − 20)
=> 30 = −k × 10 (10 − 20)
=> k = 3 10
Alternate approach:
Alternate approach: