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
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:
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:
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