GRE PAPER- III
1. DIRECTIONS: The question consists of two quantities. One in Column A and one in Column B, you are to compare the two quantities and then give the answer.
3
Each triangle is equilateral. Column A Column B Area of shaded region 6π (a) quantity in column A is greater (b) quantity in column B is greater (c) both quantities are equal (d) the relationship cannot be determined from the information given Solution. (c) Triangles are equilateral. Angles at centre are 60° each. Sum of remaining angles = 360° – (60° + 60°) = 240°.
θ 240 πr 2 = × π × (9) = 6π sq. units 360 360° Column A : Required area =
2.
DIRECTIONS: The question consists of two quantities. One in Column A and one in Column B, you are to compare the two quantities and then give the answer. Column A Column B
3 1 .2
2 2
+
5 2 .3
2 2
+
7 3 .4
2 2
15 16
(a) (b) (c) (d)
quantity in column A is greater quantity in column B is greater both quantities are equal the relationship cannot be determined from the information given
3 + 5 2 2.32 + 7 32.4 2 1 ⎞ ⎛ 1 1 ⎞ ⎛ 1 1 ⎞ ⎛1 ⎟ =⎜ − ⎟+⎜ − ⎟+⎜ − ⎜ 2 ⎟ ⎜ ⎟ ⎜ ⎟ 2 2 ⎠ ⎝ 2 2 32 ⎠ ⎝ 3 2 4 2 ⎠ ⎝1
Solution.
2 2 (c) Column A : 1 .2
= 1−
1 42
= 1−
1 15 = . 16 16
3.
PQRS is a square. SR is a tangent (at point S) to the circle with centre O and TR=OS. Then, the ratio of area of the circle to the area of the square is
P
S O T
Q
(a) π/ 3 (e) π/ 9 (b) 11 / 7
R
(c) 3 / π (d) 7 / 11
2 2 2 2 2 2 Solution. (a) In ΔSOR , a + r = (2r ) = 4r ⇒ a = 3r or a = 3r
P a Q
∴ 4.
22 2 ×r π Area of circle 22 = 7 2 = = Area of square ( 3r) 7×3 3
a S a
r r T
O
r a R
The average number of printing errors per page in a book of 512 pages is 4. If the total number of printing errors in the first 302 pages is 1208, the average number of printing errors per page in the remaining pages is (a) 0 (b) 4 (c) 840 (d) 90 (e) 9 (b) Let the average no. of printing errors per pages in the remaining (512 – 302) = 210 pages be x. ⇒ 210 x × 1208 = 512 × 4 or 210x = 2048 – 1208 = 840
or x = 840 =4 210
Solution.
5.
An exhibition tent is in the form of a cylinder surmounted by a cone. The height of the tent above the ground is 85 m and the height of the cylindrical part is 50 m. If the diameter of the base is 168 m, find the quantity of canvas required to make the tent. Allow 20% extra for folds and for stitching. (Give your answer to the nearest m2.) V
A
E
84
B
50 m
(a) 60709 (e) None of these.
m2
C (b) 60500 m2
F
D (c) 70200 m2
(d) 60509 m2
Solution.
(d) Total canvas required (without 20% extra) = Slant area of the conical part + Lateral surface area of the cylindrical part
= π . 84 (84) 2 + (35) 2 + 2π . 84 . 50
= 84π [91 + 100] = 191× 84π = 191× 84 ×
22 = 191×12 × 22 = 50424 7
with 20% extra for stitching and folds, the total canvas required = 50424 + 10084.9 = 60508.8 m2 ≅ 60509 m2 6. DIRECTIONS: The question consists of two quantities. One in Column A and one in Column B, you are to compare the two quantities and then give the answer. 36 men can complete a piece of work in 18 days. 27 men can complete the same work in x days. Column A Column B x 22 (a) quantity in column A is greater (b) quantity in column B is greater (c) both quantities are equal (d) the relationship cannot be determined from the information given
Men 36 27 Days 18 x
18× 36 = 24 27
Solution. (a)
Column A : It's indirect proportion.
∴ : 36 :: 18 : x ⇒ 27 × x = 18× 36 ⇒ x = 27
Directions for questions 7 and 8 : The following graph gives Sales, Expenses and Capital of a company for a period of five years from 1994 to 1998. Read the graph and answer the question that follow.
900 800 700 600 In Rs crores 500 400 300 200 100 0 1994
7.
Sales Expenses Capital
1995
1996
1997
1998
Years In which year was the sales-to-expenses ratio the lowest? (a) 1994 (b) 1995 (c) 1996 (e) 1998
= 400 = 1.33 300
(d) 1997
Solution.
(d) Ratio for 1994
=
Ratio for 1995
500 5 = = 1.25 400 4
=
Ratio for 1996
=
300 3 = = .75 400 4 400 4 2 = = = .66 600 6 3 800 = 8 = 1.14
Ratio for 1997
=
Ratio for 1998 700 7 Clearly, it was lowest for 1997. 8. In which year was the ratio of profits to capital the highest? (a) 1998 (b) 1995 (c) 1996 (e) can't be determined (b) 1998 0.50 Ratio of profit to capital 1995 1
(d) 1997
Solution.
1996 –1
1997 –1
9.
DIRECTIONS: The question consists of two quantities. One in Column A and one in Column B, you are to compare the two quantities and then give the answer.
P 30° O
A
O is centre. Column A
∠OAB
B
Column B 60°
(a) (b) (c) (d)
quantity in column A is greater quantity in column B is greater both quantities are equal the relationship cannot be determined from the information given
∠AOB = 2∠APB = 2 × 30° = 60°.
Solution. (c)
Q OA = OB ∴ ∠OAB= ∠OBA 1 ∠OAB = (180° − 60°) = 60°. 2 Column A :
10.
DIRECTIONS: The question consists of two quantities. One in Column A and one in Column B, you are to compare the two quantities and then give the answer. a < b and c < d Column A Column B a+c b+d (a) quantity in column A is greater (b) quantity in column B is greater (c) both quantities are equal (d) the relationship cannot be determined from the information given a < b, c < d
Solution. (b)
11.
From a pack of 52 cards, one card is drawn at random. What is the probability that the card drawn is a ten or a spade? (a)
4 13
(b)
1 4
(c)
1 13
(d)
1 26
Solution.
(a) Here, n(S) = 52 There are 13 spades (including one ten) and there are 3 more tens. Let E = event of getting a ten or a spade. Then, n(E) = (13 + 3) = 16
P(E ) =
. 12.
n (E) 16 4 = = n (S) 52 13
A ship 55 km from a shore springs a leak which admits 2 tonnes of water in 6 min. 80 tonnes would suffice to sink it, but the pumps can throw out 12 tonnes an hour. The average rate of sailing for the ship to just reach the shore before sinking is... (a) 6.5 kmph. (b) 6 kmph. (c) 5.5 kmph (d) 8 kmph. (e) 4 kmph.
80 80 = = 600 minutes ⎛ 2 12 ⎞ ⎛ 1 1 ⎞ ⎜ − ⎟ ⎜ − ⎟ (c) Time required = amount of water / accumulation rate = ⎝ 6 60 ⎠ ⎝ 3 5 ⎠
Solution.
Average sailing rate = 55 km/600 min = 55 km /10 hr = 5.5 kmph 13. DIRECTIONS: The question consists of two quantities. One in Column A and one in Column B, you are to compare the two quantities and then give the answer. If a person walks at 14 km/hr instead of 10 km/hr, he would have walked 20 km more. Column A Column B Actual distance travelled by him 45 km (a) quantity in column A is greater (b) quantity in column B is greater (c) both quantities are equal (d) the relationship cannot be determined from the information given
d d + 20 = ⇒ 14d = 10d + 200 ⇒ 4d = 200 ⇒ d = 50 km. 10 14
Solution. (a) Let the actual distance = d
∴
14.
In the figure below, AB,CD,XY are parallel to each other. If AB = 3 cm, CD = 5 cm, XY will be equal to
D
B
X
A
(a) 2 cm (e) 1.94 cm Solution. (b) 1.25 cm
Y
(c) 1.75 cm
C
(d) 1.875 cm
(d) In ΔABC, AB/XY=CA/CY or XY=3CY/CA ...(i) & In ΔADC, DC/XY=CA/AY or XY= 5AY/CA ...(ii) From (i) and (ii), we get 3CY = 5AY or CY/AY =5/3 or (CY+AY)/AY = 8/3 or AC/AY=8/3
From (ii), XY = 5 AY/AC = 5 x 3/8 = 15/8 = 1.875 15. DIRECTIONS: The question consists of two quantities. One in Column A and one in Column B, you are to compare the two quantities and then give the answer. Column A Column B The product of even integers from –4 to 4 The product of odd integers from –3 to 3 (a) quantity in column A is greater (b) quantity in column B is greater (c) both quantities are equal (d) the relationship cannot be determined from the information given Column A : Therefore product = 0. Column B : Integers are –3, –1, 1, 3. Therefore product = 9. Integers are –4, –2, 0, 2, 4.
Solution. (b)
16.
DIRECTIONS: The question consists of two quantities. One in Column A and one in Column B, you are to compare the two quantities and then give the answer.
x2 −1 =4 x +1
Column A Column B x 4 (a) quantity in column A is greater (b) quantity in column B is greater (c) both quantities are equal (d) the relationship cannot be determined from the information given
x 2 −1 = 4 ⇒ x 2 − 1 = 4( x + 1) ⇒ x 2 − 1 = 4x + 4 ⇒ x 2 − 4x − 5 = 0 (d) x + 1
⇒ x 2 − 5x + x − 5 = 0 ⇒ ( x − 5)(x + 1) = 0 ⇒ x = 5 or − 1.
Solution.
17.
Which of the following must also be a cyclic quadrilateral? I. a rectangle II. a rhombus III. an isosceles trapezium IV. a kite (a) I and IV (b) I and III (c) II and IV (e) None of these
(d) III and IV
Solution.
(b) In a cyclic quadrilateral, the sum of opposite pairs of angles is 180º, which is the case in both a rectangle and an isosceles trapezium.
18.
DIRECTIONS: The question consists of two quantities. One in Column A and one in Column B, you are to compare the two quantities and then give the answer. n is a positive integer. Column A Column B (–5)n (–5)n + 1 (a) quantity in column A is greater (b) quantity in column B is greater (c) both quantities are equal (d) the relationship cannot be determined from the information given
Solution. (d) Since, we don't know whether n is even or odd. ∴ We can't determine which column is greater than other.
x −1
19.
If f (x) = x + 1 , then find the value of f (f(x)).
1
(a) x − 1 (e) None of these
f (x) = x −1 x +1
(b)
1 x +1
1
(c) – x
(d)
1 x
Solution.
(c)
x −1 −1 x −1 − x −1 − 2 −1 x +1 = = = ⇒ f (f ( x )) = x −1 x − 1 + x + 1 2x x +1 x +1
20.
DIRECTIONS: The question consists of two quantities. One in Column A and one in Column B, you are to compare the two quantities and then give the answer.
D 5 90° A 12
C
B
Column A Column B BD 13 (a) quantity in column A is greater (b) quantity in column B is greater (c) both quantities are equal (d) the relationship cannot be determined from the information given Solution. (c) 21.
2 2 Column A : BD = 12 + 5 = 169 = 13 = Column B
In a party every person shakes hands with every other person. If there were a total of 210 handshakes in the party, find the no. of persons who were present in the party. (a) 20 (b) 21 (c) 40 (d) 42 (e) 30 (b) nC2 = 210 or n(n – 1) = 420 or n2 – n – 420 = 0
⇒ n= 1 ± 1 + (420) 2 2 = 1 ± 41 = 21, − 20 2
Solution.
Hence, the no. of person in the party = 21 22. DIRECTIONS: The question consists of two quantities. One in Column A and one in Column B, you are to compare the two quantities and then give the answer. Column A Column B
54 ÷ 66 ÷ 33
3 121
(a) (b) (c) (d)
quantity in column A is greater quantity in column B is greater both quantities are equal the relationship cannot be determined from the information given
54 ÷ 66 ÷ 33 = 54 54 1 3 ÷ 33 = × = 66 66 33 121
Solution. (c) Column A :
23.
DIRECTIONS: The question consists of two quantities. One in Column A and one in Column B, you are to compare the two quantities and then give the answer. Two dice are tossed Column A Column B The probability of getting a total more than 7 The probability of getting the total score is a prime number. (a) quantity in column A is greater (b) quantity in column B is greater (c) both quantities are equal (d) the relationship cannot be determined from the information given n (S) = 62 = 36. E = {(2, 6), (3, 6), (4, 6), (5, 6), (6, 6), (3, 5), (4, 4), (4, 5), (5, 3), (5, 4), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
15 5 = . ∴ Required probability = 36 12
Solution. (c) Column A :
Column B :
n (S) = 62 = 36 E = {(1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3), (5, 2), (5, 6), (6, 1), (6, 5)} ∴ Required probability =
n (E ) 15 5 = = n (S) 36 12
.
Directions for question 24 and 25 : Study the following graph carefully and answer the questions that follow. Imports and Exports of a country (in million dollars) Export Import 40 200 Export Import 185 36
30 30 28 150
20 11 10 6 45 0 22 1990 1991 1992 Years 1993 1994 55 11
100
50
0
24.
The export in which years was 50% of the average exports during the given years? (a) 1991 (b) 1992 (c) 1993 (d) 1994 (e) 1990
⎛ 6 + 30 + 36 + 28 + 11 ⎞ ⎜ ⎟ 5 ⎠ million dollars = 22.2 million dollars (d) Average export = ⎝
Solution.
50% of average export = 11 million dollars (approx.) This happens in the year 1994. 25. The imports in which year was nearer to the total value of imports and exports in 1994? (a) 1990 (b) 1990 & 1991 (c) 1992 (d) 1991 & 1992 (e) None of these
Solution.
(a) (Import + Export) in 1994 = (11 + 11) million dollars = 22 million dollars = Import in 1990.
1 6 h 4 7 3 4
26.
A man walks a certain distance and rides back in take to ride both ways?
4
. He can walk both ways in
4 3 4
h. How long it would
(a) 5 hours (e)
1 5 2
(b)
1 2
hours
(c)
hours
(d) 6 hours
hours
Solution.
(c)
t 2 x + t 2 y = 2( t x + t y )
t walk both + t ride both = 2( t walk + t ride )
1 3 1 3 ⎛1 3⎞ t ride both = 12 − 7 = (12 − 7) + ⎜ − ⎟ = 5 − = 4 2 4 4 4 ⎝2 4⎠
27. DIRECTIONS: The question consists of two quantities. One in Column A and one in Column B, you are to compare the two quantities and then give the answer. Column A Column B The area of a circle of radius 5 The area of a square of side 5 (a) quantity in column A is greater (b) quantity in column B is greater (c) both quantities are equal (d) the relationship cannot be determined from the information given
π(5) 2 = 25π, where π = 22 7
Solution. (a) Column A : Area of circle = Column B : Area of square = (5)2 = 25. 28.
DIRECTIONS: The question consists of two quantities. One in Column A and one in Column B, you are to compare the two quantities and then give the answer. Column A Column B The average of odd numbers upto 100 49 (a) quantity in column A is greater (b) quantity in column B is greater (c) both quantities are equal (d) the relationship cannot be determined from the information given Column A : S = 1 + 3 + 5 + ........... + 99 Total no. of terms = 50
∴S =
Solution. (a)
Numbers are 1, 3, 5, ..........., 99.
50 [ 2 × 1 + (50 − 1) 2] = 25[2 + 49 × 2] = 25[ 2 + 98] = 2500 2
1 + 3 + 5 + ........+ 99 2500 = = 50 50 50 Now, average =