Answer: Option A
Solution: (13056 Ã· 32)(513 Ã· 19) = ? + 95 \( \sqrt{1104 + 40 Ã— 30} \)
408 x 27 = ? + 95 \( \sqrt{1104 + 200} \)
11016 = ? + 95 \( \sqrt{2304} \)
? = 11016 – 95 x 48
? = 11016 – 4560
? = 6456
2. \( \sqrt{29 Ã— 1240 Ã· 31 + 173 Ã— 5} \) Ã— 36 =\( ?^2 \) + 29 Ã— 16
Answer: Option E
Solution: \( \sqrt{29 Ã— 1240 Ã· 31 + 173 Ã— 5} \) Ã— 36 =\( ?^2 \) + 29 Ã— 16
\( \sqrt{29 Ã— 40 + 865} \) Ã— 36 = \( ?^2 \) + 464
\( \sqrt{1160 + 865} \) Ã— 36 = \( ?^2 \) + 464
\( ?^2 \) = 45 x 36 – 464
\( ?^2 \) = 1620 – 464
\( ?^2 \) = 1156
\( ?^2 \) = \( \sqrt {1156}\)
\( ?^2 \) = 34
3. (16980 – 9910 – 4648) Ã· 14 = ? + (16650 Ã· 111)
Answer: Option A
Solution: (16980 – 9910 – 4648) Ã· 14 = ? + (16650 Ã· 111)
(16980 – 14558) Ã· 14 ? + 150
2422 Ã· 14 ? + 150
? + 150 = 173
? = 173 – 150 = 23
4. 73% of 4936 + 24% of 5612 – 55% of 3848 = ?
Answer: Option A
Solution: 73% of 4936 + 24% of 5612 – 55% of 3848 = ?
\( \frac{73} {100}\) x 4936 + \( \frac{24} {100}\) x 5612 – \( \frac{55} {100} \) x 3848
? = 73 x 49.36 + 24 x 56.12 – 55 x 38.48
? = 3603.28 + 1346.88 – 2116.4
? = 4950.16 – 2116.4 = 2833.76
5. \( \frac{8} {13}\) – \( \frac{7} {26}\) x \( ?^2 \) = 234
Answer: Option B
Solution: \( \frac{8} {13}\) – \( \frac{7} {26}\) x \( ?^2 \) = 234
\( \frac{16 âˆ’ 7} {26}\) \( ?^2 \) = 234
\( \frac{9} {26}\)\( ?^2 \) = 234
\( ?^2 \) = 676
? = \( \sqrt{676}\)= 26
6. \( \frac{5} {17}\) of \( \frac{51} {81} \) of 54% of 180 = \( \sqrt{?}\) – \( 9^2 \)
Answer: Option C
Solution: \( \frac{5} {17}\) of \( \frac{51} {81} \) of 54% of 180 = \( \sqrt{?}\) – \( 9^2 \)
\( \frac{5} {17}\) x \( \frac{51} {81} \) x 54 x 180 = \( \sqrt{?}\) – \( 9^2 \)
\( \frac{5} {27}\) 54 x 1.8 = \( \sqrt{?}\) – 81
18 = \( \sqrt{?}\) – 81
? = \( \sqrt{99}\) = 9801
7. 87 \( \frac{3} {8}\) + 54 \( \frac{5} {24} \) – 110 \( \frac{7} {60} \) = ?
Answer: Option A
Solution: 87 \( \frac{3} {8}\) + 54 \( \frac{5} {24} \) – 110 \( \frac{7} {60} \) = ?
(87 + 54)\( \frac{3} {8}\) + \( \frac{5} {24} \)– 110 \( \frac{7} {60} \)
(141)\( \frac{7} {12}\) – 110 \( \frac{7} {60} \)
? = (141 – 110) \( \frac{7} {12} \) – \( \frac{7} {60} \)
? = 31\( \frac{28} {60}\)
? = 31\( \frac{7} {15} \)
8. 12450 – 8421 + 5641 – 4378 = ? + 24 x (36% of 275)
Answer: Option B
Solution: 12450 – 8421 + 5641 – 4378 = ? + 24 x (36% of 275)
18091 – 12799 = ? + 24 x (36% of 275)
? + 24 x (36% of 275) = 5292
? + 24 x \( \frac{36} {100} \) x 275 = 5292
? + 24 x 99 = 5292
? = 5292 – 2376
? = 2916
9. 3\( \sqrt{?}\) = (1849 – (2112 x 25 Ã· 33)) – 240
Answer: Option D
Solution: 3\( \sqrt{?}\) = (1849 – (2112 x 25 Ã· 33)) – 240
3\( \sqrt{?}\)= (1849 – 3\( \sqrt{?}\) – (64 x 25)) – 240
3\( \sqrt{?}\)= (1849 – 3\( \sqrt{?}\) – (1600) – 240
3\( \sqrt{?}\)= 249 – 240
3\( \sqrt{?}\)= 9
? = \( 9^3\) = 729
10. \( (16)^4 \) x \( (32)^2 \) Ã· \( (8)^5 \) = \( (2)^{P^3+3} \)
Answer: Option E
Solution: \( (16)^4 \) x \( (32)^2 \) Ã· \( (8)^5 \) = \( (2)^{P^3+3} \)
\( 2^{4 Ã— 4} \)
\( (2)^{4 Ã— 4} \) x \( (2)^{2 Ã— 5} \) Ã· \( (2)^{5 Ã— 3} \) = \( (2)^{P^3+3} \)
\( (2)^{16} \) Ã— \( (2)^{10} \) Ã· \( (2)^{15} \) = \( (2)^{P^3+3} \)
\( (2)^{16 – 5}\) = \( (2)^{P^3+3} \)
\( (?){^3} \) + 3 = 11
\( (?){^3} \) = 11 – 3 = 8
? = 3\( \sqrt{8} \) = 2
Answer: Option C
Solution: (32% of 1699.98 + 71.93% of 225.10 = ? + 19% of 1101
\( \frac{32} {100} \) x \( \frac{72} {100} \)x 225 â‰ˆ ? + \( \frac{19} {100}\) x 1101
? + 19 x 11.01 = 32 x 17 + 9 x 18
? + 209 â‰ˆ 544 + 162
? = 706 – 209 = 497
2. 27.04 x 116.95 Ã· 12.98 + \( (87.01)^2 \) = ? + 7312
Answer: Option B
Solution: 27.04 x 116.95 Ã· 12.98 + \( (87.01)^2 \) = ? + 7312
27 x 117 Ã· 13 + \( (87.01)^2 \) = ? + 7312
? + 7312 = 27 x 9 + 7569
? + 7312 = 27 x 9 + 7569
? = 257 + 243
? = 500
3. \( \frac{1} {43} \) of 4817 + (82.01 x 21) = ? + (37272 Ã· 24.02)
Answer: Option D
Solution: \( \frac{1} {43} \) of 4817 + (82.01 x 21) = ? + (37272 Ã· 24.02)
\( \frac{1} {43} \) 4817 + (82 x 21) â‰ˆ ? + (37272 Ã· 24)
? + 150
? + 1553 = 112. 02 + 1722
? â‰ˆ 169 + 112
? = 281
4. 48% of 6410 + 71% of 8108 = ? \( – (45.02)^2 \)
Answer: Option A
Solution: 48% of 6410 + 71% of 8108 = ? \( – (45.02)^2 \)
\( \frac{48} {100}\) x 6410 + \( \frac{24} {100}\) x 8108 â‰ˆ ? \( – (45)^2 \)
– 2025 = 48 x 64.10 + 71 x 81.08
– 2025 â‰ˆ 3072 + 5751
? = 8823 + 2025
? = 10848
5. 27.01 of \( \frac{5} {81}\) of 8700 + \( \sqrt{27.02 Ã— 19 + 9 Ã— 24} \) = \( ?^2 \) – 24
Answer: Option C
Solution: 27.01 of \( \frac{5} {81}\) of 8700 + \( \sqrt{27.02 Ã— 19 + 9 Ã— 24} \) = \( ?^2 \) – 24
\( \frac{27} {100}\) \(\frac{5} {81} \) x 8700 + \( \sqrt{27 Ã— 19 + 9 Ã— 24} \) = \( ?^2 \) – 24
\( ?^2 \) – 24 = 5 x 29 + \( \sqrt{513 + 216} \)
\( ?^2 \) – 24 = 145 + \( \sqrt{729} \)
\( ?^2 \) = 145 + 27 + 24 = 196
\( ?^2 \) = \( \sqrt{196} \)
= 14
Number Series:
Directions for Questions (6 â€“ 10): What will come at the place of question mark (?) in the following number series?
6. 652, 211, ?, 166, 742, 117
Answer: Option B
Solution: The pattern is
652 – \( 21^2 \) = 211
211 + \( 22^2 \) = 695
695 – \( 23^2 \) = 166
166 + \( 24^2 \) = 742
742 – \( 25^2 \) = 117
7. 225, 255, 327, 383, ?, 545
Answer: Option A
Solution: The pattern is
225 + 5 x 6 = 255
255 + 6 x 7 = 327
327 + 7 x 8 = 383
383 + 8 x 9 = 455
455 + 9 x 10 = 545
8. 652, 211, ?, 166, 742, 117
Answer: Option B
Solution: The pattern is
652 – \( 21^2 \) = 211
211 + \( 22^2 \) = 695
695 – \( 23^2 \) = 166
166 + \( 24^2 \) = 742
742 – \( 25^2 \) = 117
9. 1, 17, 49, ?, 161, 24
Answer: Option E
Solution: 1 + \( 4^2 \) x 1 = 17
17 + \( 4^2 \) x 1 = 49
49 + \( 4^2 \) x 1 = 97
97 + \( 4^2 \) x 1 = 161
161 + \( 4^2 \) x 1 = 241
10. 19, 10, ?, 18, 38, 97.5
Answer: Option A
Solution: 19 x .5 +.5 = 10
10 x 1 + 1 = 11
11 x 1.5 + 1.5 = 18
18 x 2 +2 = 38
38 x 2.5 + 2.5 = 97.5
Answer: Option A
Solution: According to the question,
Perimeter of a square = 4 Ã— side
Perimeter of a square = 4 Ã— 33 = 132m
Circumference of a circle 2Ï€r
132 = 2 \( \frac{22} {7}\) Ã— r
r = 21m
Diameter of a circle = 2r
Diameter of a circle = 2 Ã— 21 = 42m
Breadth of a rectangle = 42m
Area of a rectangle = 756 \( m^2 \)
Area of a rectangle = Length of a rectangle Ã— Breadth of a rectangle
Length of a rectangle = \( \frac{756} {42}\) = 18m
Compound Interest:
2. What will be the compound interest on a sum of Rs. 44,375 at compound interest compounded annually at 12% per annum in two years?
Answer: Option C
Solution: According to the question,
P = Rs.44,375
r = 12%
t = 2 years
C.I = P({1 + \( \frac{r} {100})^t\) – 1}
C.I = 44375{(1 + \( \frac{12} {100})^2\)– 1}
C.I = 44375{( \( \frac{25 + 3} {25})^2\)– 1}
C.I = 44375{\( \frac{784 } {625}\) – 1}
C.I = 44375{\( \frac{784 – 625} {625}\)}
C.I = 44375 x \( \frac{159} {625}\)
C.I = 71 x 159 = Rs.11,289
Time and work:
3. 14 girls and 10 boys can do a piece of work in 5 days while 14 girls and 6 boys can do the same piece of work in 6 days. In how many days can 19 girls and 16 boys do the same piece of work?
Answer: Option D
Solution: 14 girls and 10 boys 1 day work
14 girls and 6 boys 1 day work
According to the question,
(14 girls + 10 boys) Ã— 5 â‰¡ (14 girls + 6 boys) Ã— 6
(70 girls + 50 boys) Ã— 5 â‰¡ (84 girls + 36 boys) Ã— 6
14 boys â‰¡ 14 girls
1 boy â‰¡ 1 girl
(19 girls + 50 boys) â‰¡ (19 girls + 16 boys) = 35 boys
M_{1} D_{1} = M_{2} D_{2}
24 x 5 = 35 D_{2}
D_{2} = \( \frac{5 Ã— 24} {35}\)
D_{2} = 3\( \frac{3} {7}\)
Percentage:
4. If the numerator of a fraction is increased by and denominator is also increased by then the value of fraction is \( \frac{4} {7}\). The original fraction is:
Answer: Option A
Solution: According to the question,
Let the original fraction be \( \frac{100 x} {100 y}\) then,
(\( \frac{100 + 450} {100 + 300}\))\( \frac{x} {y}\) = \( \frac{4} {5}\)
The original fraction
(\( \frac{550} {400}\))\( \frac{x} {y}\) = \( \frac{4} {5}\)
(\( \frac{x} {y}\)) = \( \frac{4 Ã— 8} {11 Ã— 5 }\)
(\( \frac{x} {y}\)) = \( \frac{32} {55}\)
The original fraction = \( \frac{32} {55}\)
Ratio and Proportion:
5. In a college the students in Math and Hindi classes are in the ratio 17:19. When 12 more students join Hindi class the ratio becomes 17:21. How many students are there in the Hindi class?
Answer: Option D
Solution: Let the number of students in Math and Hindi be , then According
to the question,
(\( \frac{17 x} {19 x + 12}\)) = \( \frac{17} {21}\)
357x = 323x + 204
357x – 323x – 204
34x = 204
x = 6
Number of students in Hindi class = 19x = 19 Ã— 6 =114
Pipes and Cistern:
6. Pipe J can fill a tank in 14hours while pipe K alone can fill it in 28hours and pipe L can fill a tank in 42hours. Pipe J was opened and after one hour Pipe K was opened and after two hours from start of Pipe J, Pipe L was also opened. Find the time in which the cistern is just full?
Answer: Option E
Solution: According to the question
Part of cistern filled by Pipe J in first 2 hours = \( \frac{1} {14} Ã— 2 \) = \( \frac{1} {7}\)
357x – 323x – 204
34x = 204
x = 6
Number of students in Hindi class = 19x = 19 Ã— 6 = 114
Part of cistern filled by Pipe K in one hour = \( \frac{1} {28} \)
Part of cistern filled by Pipe K in one hour = \( \frac{1} {28}\)
Remaining to the filled = 1 – \( \frac{1} {7} \) – \( \frac{1} {28} \)
Pipe J, K and L together 1hour filled = \( \frac{1} {14} \) + \( \frac{1} {28} \) + \( \frac{1} {42} \) = \( \frac{6 + 3 + 2} {84} \)
\( \frac{23} {28} \) part of cistern is filled by pipe J, K and L \( \frac{23/28} {11/84} \)
= \( \frac{23 Ã— 84} {28 Ã— 11} \)
= \( \frac{69} {11} \)
= 6\( \frac{3} {11} \)hours
Profit and Loss:
7. A shopkeeper sells one article for Rs.720 at a gain of 20% and another for Rs.640 at a loss of 20%. His total gain or loss percent is:
Answer: Option D
Solution: According to the question,
C.P. of 1st article \( \frac{720 Ã— 100} {120} \) = Rs600
C.P. of 2nd article \( \frac{640 Ã— 100} {80} \) = Rs800
Total C.P. = 600 + 800 = Rs 1400
Total S.P. = 720 + 640 = Rs 1360
Loss = 1400 – 1360 = Rs 40
Loss% = \( \frac{40 Ã— 1400} {100} \) = \( \frac{40} {14} \)
= 2.86%
Average:
8. The sum of three numbers is 195. If the ratio of the first and the second is 4:5 and that of the second and the third is 5:4, then what is the sum of first and third numbers?:
Answer: Option E
Solution: According to the question,
Let the three numbers be a, b, and c, then
a : b = 4 : 5
b : c = 5 : 4
a : b : c = 20 : 25 : 20
a = \( \frac{195 } {20 + 25 + 20} \) x 20
a = \( \frac{195 } {65} \) x 20 = 60
c = \( \frac{195 } {20 + 25 + 20} \) x 20
c = \( \frac{195 } {65} \) x 20 = 60
sum of first and third number = 60 + 60 = 120
Probability:
9. A box contains 4 blue, 8 grey and 8 white balls. If 4 balls are picked at random. What is the probability that two are grey and two are white?:
Answer: Option E
Solution: Total number of balls in the box = 4 + 8 + 8 = 20
P = \( \frac{n(E)} {n(S)} \)
n(E) = \(8_{{C}_{2}}\) x \(8_{{C}_{2}}\)
= !\( \frac{!8} {!2 Ã— !8 – 2} \) = \( \frac{8 Ã— 7 } {2} \) X \( \frac{8 Ã— 7 } {2} \)
= 28 X 28 = 784
n(S)= \(20_{{C}_{4}}\) = !\( \frac{!120} {!4 Ã— !120 – 4} \) = \( \frac{20 Ã— 19 Ã— 18 Ã— 17} {4 Ã— 3 Ã— 2 Ã— 1} \)
= 5 Ã— 3 Ã— 19 Ã— 17 = 4845
P = \( \frac{784} {4845} \)
Permutation and Combination:
10. In how many different ways can the letters of the word “HELICOPTER” be arranged so that the consonant always come together?:
Answer: Option E
Solution: In word “HELICOPTER” there are 10 letters in which consonants are H, L, C, P, T and R and vowels are E, I, O, and E. E occurs twice So it can be arranged in following ways-
= \( \frac{!6 Ã— !5} {!2} \)
= \( \frac{6 Ã— 5 Ã— 4 Ã— 3 Ã— 2 Ã— 1} {2 Ã— 1} \)
= 720 Ã— 60 = 43200
Answer: Option D
Solution: Number of male workers working in factory G in 2011
= 180 x 100 = 18000
Number of male workers working in factory G in 2010
120 x 100 = 12000
Percentage increase = \( \frac{18000 – 12000} {12000} \) x 100
= \( \frac{6000} {12000} \) x 100
= 50%
2. Number of male workers working in factory H in 2014 and 2015 together are approximately what percent of number of male workers working in factory I in 2010 and 2015 together?
Answer: Option B
Solution: Number of male workers working in factory H in 2014 and 2015 together
(180 + 260) x 100 = 44000
Number of male workers working in factory I in 2010 and 2015 together
(320 + 340) x 100 = 44000
Percentage = \( \frac{44000} {66000} \) x 100
= \( \frac{4400} {6600} \)
= 66.7% â‰ˆ 67%
3. What is the difference between number of male workers working in factory I in 2010, 2011, 2012, 2013 and 2015 together and number of male workers working in factory G, H, I and J together in 2013?
Answer: Option E
Solution: Number of male workers working in factory I in 2010, 2011, 2012, 2013 and 2015 together
= (320 + 240 + 180 + 140 + 340) x 100
= 12200
= Number of male workers working in factory G, H, I and J together in 2013
= (240 + 180 + 140 + 340) x 100
= 10000
Difference = 12200 – 10000
= 22000
4. What is the percentage decrease in number of male workers working in factory H in 2013 as compared to the previous year?
Answer: Option D
Solution: Number of male workers working in factory H in 2013
280 X 100 = 28000
Number of male workers working in factory H in 2012
320 X 100 = 32000
Percentage decrease = \( \frac{32000 – 28000} {32000} \) x 100
= \( \frac{4000} {32000} \) x 100
= \( \frac{100} {8} \)
= 12.5%
5. What is the ratio between number of male workers working in factory G and J together in 2014 to number of male workers working in factory H and I together in 2015?
Answer: Option A
Solution: Number of male workers working in factory G and J together in 2014
Number of male workers working in factory G and J together in 2014
= (300 + 280) x 100
= 58000
Number of male workers working in factory H and I together in 2015
= (260 + 340 x 100
= 60000
Ratio = \( \frac{58000} {60000} \)
= \( \frac{58} {60} \)
= 29 : 30
Correct Answer: 3
2. Answer of following (aÂ³b)^{4}is ________.
Correct Answer: 3
3. Tan(Î± + Î²) = ___________.
Correct Answer: 2
4. Path described by any moving point is classified as ___________.
Correct Answer: 4
5. The sector of a circle has radius of 21 cm and central angle 135o. Find its perimeter?
Correct Answer: 1
Solution:
Perimeter of the sector = length of the arc + 2(radius)
= (135/360 * 2 * 22/7 * 21) + 2(21)
= 49.5 + 42 = 91.5 cm
6. The middle value of an ordered array of numbers is the
Correct Answer: 3
7. Which of the following is not a measure of central tendency?
Correct Answer: 3
Solution:
Standard Deviation is measure of Dispersion or variability its not measure of central tendency.
8. What will be the cost of building a fence around a square plot with area equal to 289 sq ft, if the price per foot of building the fence is Rs. 58?
Correct Answer: 1
Solution:
Let the side of the square plot be a ft.
a^{2} = 289 => a = 17
Length of the fence = Perimeter of the plot = 4a = 68 ft.
Cost of building the fence = 68 * 58 = Rs. 3944.
9. In a triangle ABC, if angle A = 72Â° , angle B = 48Â° and c = 9 cm then Äˆ is
Correct Answer: 3
10. In a chess tournament each of six players will play every other player exactly once. How many matches will be played during the tournament?
Correct Answer: 2
Solution:
III. matches of third player with 3 players other than the first player and second player.
So total matches will be 5+4+3+2+1 = 15