1. The number of zeros at the end of the product of the expression 10 × 100 × 1000 × 10000 × ....10000000000 is :
(A) 10 (B) 100 (C) 50 (D) 55
2. Find the unit digit of the product of all the odd prime numbers.
(A) 2 (B) 3 (C) 5 (D) 7
3. A leading chocolate producing company produces ‘abc’ chocolates per hour (abc is a three digit positive number). In how many hours it will produce ‘abcabc’ chocolates ?
(A) abc (B) 101 (C)1001 (D) can’t be determined
4. Total number of factors of the expression 623
– 543– 83 is :
(A) 60 (B) 62 (C) 46 (D) can’t be determined
5. A number ‘p’ is such that it is divisible by 7 but not by 2. Another number ‘q’ is divisible by 6 but not by 5, then the following expression which necessarily be an integer is :
(A) 42 q 6 p 7 (B) 71 q 6 p 5 (C) 42 q 7 p 6 (D) none of these 6. If 223 + 233 + 243 + .. + 873 + 883 is divided by 110 then the remainder will be :
(A) 55 (B) 1 (C) 0 (D) 44
7. If (x – 5)(y + 6)(z – 8) = 1331, then the minimum value of x + y + z is :
(A) 40 (B) 3 (C) 19 (D) not unique
8. The quotient when L.C.M. is divided by the H.C.F. of a G.P. with first term ‘a’ and common ratio ‘r’ is :
(A) rn – 1 (B) rn (C) a–1rn–2 (D) (rn
– 1)
9. Number of pairs of positive integers which satisfy the equation b a 1 b 1 a = 11 where a + b 100 (A) 6 (B) 7 (C) 8 (D) 9
10. Convert (231)8 into decimal system :
(A) 163 (B) 153 (C) 123 (D) 113
IIT-JEE | AIPMT | AIEEE | OLYMPIADS | KVPY | NTSE
WORKSHOP FOR IJSO
DAILY PRACTICE PROBLEMS
SESSION-2011-12
Subject : Mathematics Topic : NUMBER SYSTEM DPP No. 01
1. If sin (A – B) = 2 1 and cos (A + B) = 2 1
, 0º < A + B 90º, then A and B are
(A) 45º, 15º (B) 60º, 30º (C) 30º, 15º (D) 45º, 30º
2. If sec A + tan A = p , then the value of sin A is
(A) 2 p 1 p 2 – 1 (B) p 1 1 – p 2 2 (C)
2
2 p – 1 2 p 1 (D) 1 – p 1 p 2 2 3. If AD = 2 1BD, then the value of sin is
(A) 2 2 x 5 y 9 x (B) 9y2–5x2 x 2 (C) 2 2 x 5 – y 3 x (D) 2 2 x 4 – y 3 x 2
4. If sin 3 = cos (– 6º) , where 3and – 6º are acute angles, find the value of
(A) 42º (B) 21º (C) 24º (D) None of these
5. sin 1 sin 1 + sin 1 sin 1 is equal to :
(A) – 2 sec (B) 2 sec (C) 2 cosec (D) 2 tan
6. sin6A + cos6 A is equal to :
(A) 1 – 3 sin2 A cos2A (B) 1 – 3sin A cos A (C) 1 + 3 sin2 A cos2A (D) 1 7. If sec x = P, cosec x = Q, then :
(A) P2 + Q2 = PQ (B) P2 + Q2 = P2Q2 (C) P2
– Q3 = P2Q2 (D) P2 + Q2 = – P2Q2 8. ABCD is a parallelogram, where AB = 6 3 cm, BC = 6 cm and ABC = 120º. The bisectors of the
angles A, B, C and D form a quadrilateral PQRS. Find the area of PQRS (in cm2)
(A) 18 3 (B) (2 3) (C)
3 36
(D) 18 3(2 3)
9. The angle of elevation of the top of a tower as observed from a point on the horizontal ground is ‘x’. If we move a distance ‘d’ towards the foot of the tower, the angle of elevation increases to ’y’, then the height of the tower is : (A) x tan – y tan y tan x tan d
(B) d(tan y + tan x) (C) d(tan y – tan x) (D)
x tan y tan y tan x tan d 10. Let be an acute angle such that sec2 +tan2 = 2. The value of (cosec2 + cot2), is
(A) 9 (B) 5 (C) 4 (D) 2
11. If x = r sin cos , y = r sin sin , z = r cos then the value of x2 + y2 + z2 is
(A) 0 (B) 1 (C) r2 (D) None
IIT-JEE | AIPMT | AIEEE | OLYMPIADS | KVPY | NTSE
WORKSHOP FOR IJSO
DAILY PRACTICE PROBLEMS
SESSION-2011-12
Subject : Mathematics Topic : TRIGONOMETRY DPP No. 02
12. Given 3 sin + 5 cos = 5, then the value of (3 cos – 5 sin )2 is equal to (A) 9 (B) 5 9 (C) 3 1 (D) 9 1
13. If tan4 + tan2 = 1, then cos4 + cos2 has the value equal to : (A) 2 1 (B) 1 (C) 4 1 (D) 2 3
14. On the level ground, the angle of elevation of the top of a tower is 30º. On moving 20 m nearer, the angle of elevation is 60º. The height of the tower
is-(A) 10 m (B) 15 m (C) 10 3m (D) 20 m
15. The shadow of a pole standing on a horizontal plane is a metre longer when the sun’s elevation is than when it is . The height of the pole of will be :
(A)
cos cos cos a m (B)
sin sin sin a m (C)
sin cos sin a m (D)
cos cos sin a m16. An aeroplane flying horizontally 1 km above the ground is observed by person on his right side at an elevation of 60º. If after 10 seconds the elevation is observed to be 30º, from the same point and in the same direction, then uniform speed per hour (in km) of the aeroplane is (neglect the height of the person for computations).
(A) 360 3 (B)
3 720
11. The expression 2 . 1 1 + 3 . 2 1 + 4 . 3 1 + ...+ ) 1 n ( n 1
for any natural number n, is : (A) always greater than 1 (B) always less than 1
(C) always equal to 1 (D) not definite
12. The digit at the 100th place in the decimal representation of 7 6
, is :
(A) 1 (B) 2 (C) 4 (D) 5
13. In how many ways can 576 be expressed as the product of two distinct factors ?
(A) 10 (B) 11 (C) 12 (D) 13
14. Find the highest power of 63 whcih can exactly divide 6336!.
(A) 2050 (B) 1054 (C) 1020 (D) 2120
15. Find the unit digit of the product of all the odd prime numbers.
1. In the given figure, calculate the measure of POR ; where O is the center of the circle.
(A) 120º (B) 140º (C) 80º (D) 60º
2. In the given figure, the value of x is :
(A) 30º (B) 40º (C) 45º (D) 60º
3. What is the value of ‘d’ in the given figure ?
(A) 150º (B) 60º (C) 105º (D) 90º
4. ABCD is a cyclic quadrilateral such that ADB = 30º and DCA = 80º. Find the value of DAB.
(A) 70º (B) 100º (C) 120º (D) 150º
5. It is given that AB and AC are the equal sides of an isosceles ABC, in which an equilateral DEF is inscribed. As shown in the figure, BFD = a and ADE = b, and FEC = c. Then :
D E A B F C a b c (A) a = 2 c b (B) b = 2 c a (C) c = 2a + 2b (D) a = 3 c b
IIT-JEE | AIPMT | AIEEE | OLYMPIADS | KVPY | NTSE
WORKSHOP FOR IJSO
DAILY PRACTICE PROBLEMS
SESSION-2011-12
Subject : Mathematics Topic : GEOMETRY DPP No. 03
6. In a right-angled triangle, the product of two sides is equal to the half of the square of the third side, i.e. hypotenuse. One of the acute angles may be
(A) 60º (B) 30º (C) 45º (D) 15º
7. In the given parallelogram ABCD, if 3(BE) = 2(DC) and the area of DQC is 36 square unit, then find the area of BQE (sq. unit) :
A E B
Q
C D
(A) 16 (B) 20 (C) 24 (D) 18
8. In the following figure, if MN || BC, MN divides the triangle into two equal parts, then the value of the ratio of MA and AB will be : M N A B C (A) 2 (B) 2 1 (C) 2 1 2 (D) 2 1 2
9. In the given triangle ABC, points P, Q and R divide the sides BC, CA and AB in the ratios 1 : 2, 3 : 2 and 3 : 2 respectively. Find the ratio of the area of quadrilateral ARPQ to the area of the triangle ABC.
A R B P Q C (A) 15 7 (B) 3 2 (C) 5 2 (D) 75 38
10. As shown in the figure, OD = 36 cm, OA = 20 cm and AB = 25 cm. Find the length of chord BC.
B C A D 36 2520 O (A) 48 cm (B) 68 cm (C) 56 cm (D) 63 cm
11. Two circles of radius r and centres O1 and O2 are moved towards one another and AB is the common chord. The line joining O1 O2 when extended meets the circumference of one of circles at P. What is the maximum possible area of APB ?
O1 O2 A B P (A) r2 (B) 3 4 r2 (C) 4 3 3 r2 (D) 2r2
12. P and Q are the points on the side BC, R and S are on the side CA, and T is on the side AB of a ABC such that P and Q trisect BC, and CR : RS : SA = 1 : 1 : 2. T bisects AB. If area of the triangle ABC = M sq. units, the area of pentagon PQRST is :
(A) 3 M (B) 4 M (C) 3 M 2 (D) 2 M
13. RM is the direct common tangent to the circles with centres C1 and C2. Points C1, P, Q, C2 K and N are six distinct points on the same straight line. The radii of both the circles are integral multiples of a cm and radius of the circle with C1 as the center is greater than that of the circle with C2 as the center. If PQ = 2cm and RM = 8 cm, then find the length of KN.
C1 C2 K Q P M R N (A) 5 9 (B) 3 cm (C) 4 11 (D) 2 cm
14. The inscribed circle of right angled triangle ABC touches the sides AB, BC and CA at D, E and F respectively. If AD = 6 cm and BE = 5 cm, then find the length of AC.
A D F E B 5 C 6 (A) 59 cm (B) 57 cm (C) 55 cm (D) 61 cm
15. In PQR, PQ = PR, S and T are points on PR and PQ respectively such that RQ = QS = ST = TP. PTS equals : P T S R Q (A) 7 (B) 7 2 (C) 7 3 (D) 7 5
16. In the figure, AB and CD are diameters of the circle. AB is perpendicular to CD and chord DF intersects AB at E. If DE = 6 units and EF = 2 units, then the area of the circle is :
C F B O A D E (A) 32 (B) 22 (C) 36 (D) 24
17. In the adjoining figure, ABCD is a parallelogram. AD is parallel to FE and FB AF is 3 2 then find GD BG . A B C D E F G (A) 5 3 (B) 8 3 (C) 2 1 (D) 5 2
18. Consider the rectangle ABCD as shown, E is a point on CD su ch that AE = 3 unit, BE = 4 unit and AE BE. The area of rectangle ABCD is :
A
D C
B
3 4
E
1. The equation 2x + x3 = 9 has :
(A) Two real roots and one imaginary roots (B) One real & one imaginary
(C) Two imaginary roots (D) Two real roots
2. Rahul and Sameer solved a quadratic equation while solving it, Rahul made a mistake in the constant term and obtained the roots as 5, – 3. Where as sameer made a mistake in the coefficient of x and obtained the roots as 1, – 3. The correct roots of quadratic equation are :
(A) 1, 3 (B) – 1, 3 (C) – 1, – 3 (D) 1, – 1
3. If f(x) = ax2 + bx + c, g(x) =
– ax2 + bx + c, where ac 0, then f(x).g(x) = 0, has : (A) at least three real roots (B) no real roots
(C) at least two real roots (D) two real roots and two imaginary roots 4. If a3 = b3 and a
b, then the sum of the roots of equation x2– (a2 + ab + b2) x + k = 0 is equal to :
(A) 0 (B) k (C) a2 (D) b2
5. The number of roots satisfying the equation 5x = x 5x is/are :
(A) 1 (B) 3 (C) 2 (D) 4
6. The possible values of the coefficient ‘a’ for which x2 + ax + 1 = 0 and x2 + x + a = 0 have at least one common root are :
(A) a = 1 & 2 (B) a = 1 & – 2 (C) a = – 1 & 2 (D) a = – 1 & – 2 7. If 4a – 5 + b = 2a + b× 2b× 2a – 4– 63, then find the sum of a and b.
(A) 2 (B) 3 (C) 4 (D) 5 8. (2x2 + 3x + 5)1/2 + (2x2 + 3x + 20)1/2 = 15, therefore x is : (A) 3 8 (B) 5 14 (C) 4 & 2 11 (D) 4 9. Find the sum of the series
) 7 3 ( 1 + (7 11) 1 + (11 15) 1 + ... (A) 3 1 (B) 6 1 (C) 12 1 (D) 24 1
10. The roots of the equation 12x2 + mx + 5 = 0 will be in the ratio 3 : 2, if m equals :
(A) 12 1 (B) 10 12 5 (C) 12 10 5 (D) ±5 10
IIT-JEE | AIPMT | AIEEE | OLYMPIADS | KVPY | NTSE
W O R K S H O P F O R I J S O
DAILY PRACTICE PROBLEMS
S E S S I O N - 2 0 1 1 - 1 2
Subject : Mathematics Topic : ALGEBRA AND SETS DPP No. 04
11. What is the maximum possible value of
y x
for which (x – 2)2 = 9 and (y – 3)2 = 25 ? (A) 2 1 (B) 8 5 (C) 8 1 (D) 2 5
12. The set {x : (x – 3)(x – 5) > 0} is equal to :
(A) {x : 3 < x < 5} (B) {x : x < 3} {x : x < 5} (C) {x : x < 3} (x : x > 5} (D) none of these
13. If A is the set of all integral multiples of 3 and B is the set of all integral multiples of 5, then A B is the set of all integral multiples of :
(A) 3 + 5 (B) 5 – 3 (C) GCD (3, 5) (D) LCM (3, 5)
14. If A has 3 elements and B has 6 elements, then the minimum number of elements in A B is :
(A) 3 (B) 6 (C) 9 (D) 18
15. If the sets A and B are defined as A =
, 0 x R x 1 y ; y , x , B =
x,y
;yx,xR
, then(A) ABA (B) ABB (C) AB (D) None of these
16. Out of 800 boys in a school, 224 played cricket, 240 played hockey and 336 played basketball. Of the total, 64 played both basketball and hockey ; 80 played cricket and basketball and 40 played cricket and hockey 24 player all the three games. The number of boys who did not play any game is
(A) 128 (B) 216 (C) 240 (D) 160
17. If a2 + b2 + c2 = 1, then which of the following cannot be a value of (ab + bc + ca) ?
(A) 0 (B) 2 1 (C) 4 1 (D) – 1 18. 3 1 2 3 1 3 3 x y ) xy ( y x y x (A) x + y (B) y – x (C) x 1 – y 1 (D) x 1 + y 1 19. If a1/3 + b1/3 + c1/3 = 0, then :
(A) a + b + c = 0 (B) a + b + c = 3abc (C) a3 + b3 + c3 = 0 (D) (a + b + c)3 = 27abc 20. In a group of 500 people, 200 can speak Hindi alonewhile only 125 can speak English alone. The number
of people who can speak both Hindi and English is :
1. A cow is tied by a rope to one of the vertices of a square of side 14cm. The length of the rope is 7 cm. What percentage of the field is grazed by the cow ?
(A) 15% (B) 25% (C) 18% (D) 19.6%
2. A solid sphere is cut into 16 identical with 5 cuts. What is the percentage increase in the combined total surface area of all the pieces over that of the original sphere ?
(A) 350% (B) 150% (C) 200% (D) 250%
3. There is a cage in hemisphere-shape in which a canary sleeps at the centre of the base. It wakes up, flies to the top-most point in the cage, then in a straight line to the cage door at the intersection of the curved surface and the base. It covers a total distance of 241 yards. Find out the approximate radius of the hemispheric cage.
(A)120.5 yards (B) 140 yards (C) 50 yards (D) 100 yards
4. If the perimeter of a rectangle is p and its diagonal is d, then the difference between the length and width of the rectagle is (A) 2 p d 8 2 2 (B) 2 p d 8 2 2 (C) 2 p d 6 2 2 (D) 4 p d 8 2 2
5. On a square kilometre land, 2 cm of rain has fallen. Assuming that 50% fo the raindrops could have been collected and contained in a pool having a (100 × 10) m2 base, find to what level the water level in the pool has increased. (You have to assume that base of the pool is horizontal.
(A) 9.86 m (B) 15 m (C) 10 m (D) 20 m
6. A large solid sphere is melted and moulded to form identical right circular cones wih base radius and height same as the radius of the sphere. One of these cones is melted and moulded to from a smaller solid sphere. What is the ratio of the surface area of the smaller sphere to the surface area of the larger sphere ?
(A) 1 : 34/3 (B) 1 : 23/2 (C) 1 : 33/2 (D) 1 : 24/3
7. A closed wooden box measures externally 9 cm long, 7 cm broad and 6 cm high. If the thickness of the wood is half a centimetre, find the capacity of the box.
(A) 278 cm3 (B) 215 cm3 (C) 224 cm3 (D) 240 cm3
8. An equilateral triangle of side 6 cm is cut into smaller equilateral triangle of side 2 cm. What is the greatest number of the smaller triangles that can be formed ?
(A) 9 (B) 6 (C) 12 (D) 15
9. A goat is tied at A, one of the vertices of a building which is in the shape of an equilateral triangular prism. The length of each side of the building is 6 cm. The length of the rope to which the goat is tied with is 12m. There is a huge lawn on the unshaded part of the compound. What is the area of the lawn that the goat can graze on ? (Figure shows the view from the top.)
A
(A) 144 (B) 128 (C) 84 (D) 60
IIT-JEE | AIPMT | AIEEE | OLYMPIADS | KVPY | NTSE
WORKSHOP FOR IJSO
DAILY PRACTICE PROBLEMS
SESSION-2011-12
Subject : Mathematics Topic : MENSURATION DPP No. 05
10. The diameter of a circle is 10 cm. The radius of this circle is taken as a diameter and another circle is drawn. A third circle is drawn with the radius of the second circle as its diameter. This process is repeated n times until the diameter of the nth circle is less than 0.01 cm. What is the value of n ?
(A) 9 (B) 10 (C) 11 (D) 12
11. Let A0 A1 A2 A3 A4 A5 be a regular hexagon inscribed in a circle of unit radius.Then the product of the lengths of the line segments A0 A1, A0 A2 & A0 A4 is:
(A) 4 3 (B) 3 3 (C) 3 (D) 2 3 3
12
.
Two cylinders of same volume have their heights in the ratio 1 : 3, find the ratio of their radii –(A) 3 : 1 (B) 2 : 1 (C) 5 : 2 (D) 2 : 5
13. If the radii of the ends of a bucket, 45 cm high are 28 cm, and 7 cm, determine it's surface area. (A) 555 cm2 (B) 545.50 cm2 (C) 561.49 cm2 (D) 567.49 cm2
14. The height of a cone is 30 cm. A small cone is cut off at the top by a plane parallel to the base. If its volume be
27 1
of the volume of the given cone, then the height above the base at which section has been made is–
(A) 10 cm (B) 15 cm (C) 20 cm (D) 25 cm
15. If h be the height and the semi vertical angle of a right circular cone, then its volume is given by -(A) 3 1 h3 tan2 (B) 3 1 h2 tan2 (C) 3 1 h2 tan3 (D) 3 1 h3 tan3
16. The slant height of a cone is increased by P%. If radius remains same, the curved surface area is increased by –
(A) P% (B) P2% (C) 2 P% (D) None
17. A hollow spherical ball whose inner radius is 4 cm is full of water. Half of the water is transferred to a conical cup and it completely filled the cup. If the height of the cup is 2 cm, then the radius of the base of cone in cm is :
(A) 4 (B) 8 (C) 8 (D) 16
18. A square and an equilateral triangle are inscribed incircle of radius 2007 cm. The ratio of the squares of their sides is
(A) 2 : 3 (B) 3 : 2 (C) 3 : 4 (D) 4 : 3
19. In the diagram ABC is right angled at C. Also M, N and P are the mid points of sides BC, AC and AB, respectively. If the area of APN is 2 sq. cm, then the area of ABC, in sq. cm is :
(A) 8 (B) 12 (C) 16 (D) 4
20. Two circles occupy the positions shown with respect to the two squares. The two circles are each inscribed in their square and the small square is inscribed in the large circle. What is the difference in the areas of the two shaded regions if the small square has sides which measure 8 cm ?
