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1. In the first 10 overs of a cricket game, the run rate was only 3.2. What should be the run rate in the remaining 40 overs to reach the target of 282 runs?
6.25
6.5
6.75
7
Answer: Option a
Explanation:
Required run rate ( (282-(3.2 ×10))/40 )= 250/40=6.25
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2. A grocer has a sale of Rs. 6435, Rs. 6927, Rs. 6855, Rs. 7230 and Rs. 6562 for 5 consecutive months. How much sale must he have in the sixth month so that he gets an average sale of Rs. 6500?
Rs. 4991
Rs. 5991
Rs. 6001
Rs. 6991
Answer: Option a
Explanation:
Total sale for 5 months = Rs. (6435 + 6927 + 6855 + 7230 + 6562) = Rs. 34009.
Required sale = Rs. [ (6500 x 6) – 34009 ]
= Rs. (39000 – 34009)
= Rs. 4991.
3. Which of the following fraction is the least ?
A. 12/13
B. 14/19
C. 17/21
D. 25/26
Option 2 : 14/19
SOLUTION:
Calculation:
Option 1:
12/13 = 0.9230
Option 2:
14/19 = 0.7368
Option 3:
17/21 = 0.8095
Option 4:
25/26 = 0.9615
∴ The fraction 14/19 is the smallest.
The correct option is 2 i.e. 14/19
4. Simplify the following.
a. 4.5
b. 7.25
c. 1.875
d. 5.75
Option c : 1.875
SOLUTION:
Calculation:
Let us convert all the mixed fractions into improper fractions. Then, the question will look like:
15/2 ÷ {17/4 – 1/2 × (5/2 – 5/4 – 3/4)}
⇒ 15/2 ÷ {17/4 – 1/2 × 1/2}
⇒ 15/2 ÷ {17/4 – 1/4}
⇒ 15/2 ÷ 4
⇒ 15/2 × 1/4 = 15/8 = 1.875
∴ 1.875
5. When a sum is lent at 15% per annum then the difference between C.I. and S.I of 2 years is Rs. 27. Find the interest if the same sum is compounded at 20% half yearly for 1 year.
Rs.272
Rs. 250
Rs. 252
Rs. 324
Option 3 : Rs. 252
Given:
The difference between C.I. and S.I of 2 years is Rs. 27
Formula used:
Difference between C.I. and S.I of 2 years = P (r/100)2
Calculation:
Using formula get,
27 = P (15/100)²
⇒ P = Rs. 1200
Now, Rs. 1200 compounded at 20% half yearly for 1 year
⇒ P = 1200, R% = 20/2 = 10%, t = 2
A = P × (1+ r/100)^t
A = 1200 × (1+ 10/100)2 = 1200 × (11/10)2 = Rs. 1452
C.I. = A – P = 1452 – 1200 = Rs. 252
∴ Required interest is Rs. 252.
6. Amal and Bimal invested some amount in a bank at an interest rate of 10%. Amal selected a plan in which interest amount calculated annually in simple method. In Bimal’s plan interest calculated annually by the compound interest method. If both received the same interest amount after 2 years, what will be the ratio of investments made by Amal and Bimal?
21 : 20
11 : 10
10 : 11
20 : 21
Option 1 : 21 : 20
Calculation:
Let m and n are the investments by Amal and Bimal
Interest amount received by amal after 2 years = m × (r/100) × t
⇒ I = m × (10/100) × 2
⇒ I = 0.2m —(1)
Interest amount received by Bimal after 2 years = n ( 1 + (r/100))t – n
⇒ I = n × ( 1 + (10/100))2 – n
⇒ I = 1.12n – n
⇒ I = 0.21n —(2)
(1) = (2)
⇒ 0.2m = 0.21n
⇒ m/n = 21/20
7. The probability that cannot exist among the following:
(a) ⅔
(b) -1.5
(c) 15%
(d) 0.7
Answer: (b) -1.5
Explanation: The probability lies between 0 and 1. Hence, it cannot be negative.
8. If P(E) = 0.07, then what is the probability of ‘not E’?
(a) 0.93
(b) 0.95
(c) 0.89
(d) 0.90
Answer: (a) 0.93
Explanation: P(E) + P(not E) = 1
Since, P(E) = 0.05
So, P(not E) = 1 – P(E)
Or, P(not E) = 1 – 0.07
∴ P(not E) = 0.93
9. A bag has 3 red balls and 5 green balls. If we take a ball from the bag, then what is the probability of getting red balls only?
(a) 3
(b) 8
(c) ⅜
(d) 8/3
Answer: (c) ⅜
Explanation: Number of red balls = 3
Number of green balls = 5
Total balls in bag = 3+5 = 8
Probability of getting red balls = number of red balls/total number of balls
= ⅜
10. A alone can do a piece of work in 6 days and B alone in 8 days. A and B undertook to do it for Rs. 3200. With the help of C, they completed the work in 3 days. How much is to be paid to C?
Rs. 375
Rs. 400
Rs. 600
Rs. 800
Answer: Option b
Explanation:
C’s 1 day’s work = 1/3-[1/6+ 1/8]= 1/3- 7/24= 1/24
A’s wages : B’s wages : C’s wages = 1/6 ∶ 1/8 ∶ 1/24=4∶3∶1
C’s share (for 3 days) = Rs. {3 × 1/24 ×3200}=Rs.400
11. If 40% of 40% of x is 40 then ‘x’ is
100
250
400
None of these
Option 2 : 250
Calculation:
40% of 40% of x is 40
12. What will come in the place of the question mark (?) in the following question?
(15% of 2500 – 25) ÷ 20% of 35 + 50% of 200 = ?
125
165
185
215
150
Option 5 : 150
Given :
⇒ (15% of 2500 – 25) ÷ 20% of 35 + 50% of 200 = ?
⇒ (15 × 2500/100 – 25) ÷ 20 × 35/100 + 50 × 200/100 = ?
⇒ (375 – 25) ÷ 7 + 100 = ?
⇒ 350 ÷ 7 + 100 = ?
⇒ 50 + 100
⇒ ? = 150
Hence, 150 is the correct answer.
13. If 885 is 15% of a given number. So, what is the number?
5900
5750
5850
5100
Option 1 : 5900
Calculation:
15% of x = 885
15x/100=885
X = (885×100)/15
X =5900
Therefore, the required number is 5900.
14. The sum of two numbers is 384 HCF of the numbers is 48 the difference of the number is
100
192
288
336
Not Attempted
Option 3 : 288
Calculation:
Let the numbers be 48p and 48q , where p and q are co-primes.
According to the question
⇒ 48p + 48q = 384
⇒ 48(p + q) = 384
⇒ p + q = 8 ….(1)
Possible and acceptable pairs of p and q satisfying this condition is: (1, 7) and (3, 5).
So, the numbers = (48 × 1) = 48 and (48 × 7) = 336
And, (48 × 3) = 144 and (48 × 5) = 220.
So, the required difference = (336 – 48) = 288
∴ The require answer is 288.
15. Find the LCM of 315, 840 and 1680.
3420
4580
4950
5040
5120
Option 4 : 5040
Calculation:
According to the question
⇒ 315 = 3 × 3 × 7 × 5 = 32 × 7 × 5
⇒ 840 = 2 × 2 × 2 × 3 × 5 × 7 = 23 × 3 × 5 × 7
⇒ 1680 = 2 × 2 × 2 × 2 × 3 × 5 × 7 = 24 × 3 × 5 × 7
The LCM of (315, 840, 1680) will be
⇒ 24 × 32 × 5 × 5 = 5040
∴ The required result will be 5040.
16. If GCD and LCM of two numbers are 29 and 174 respectively and the numbers are in the ratio 1 : 6, then what is the largest number amongst the two numbers?
152
174
166
146
Not Attempted
Option 2 : 174
Calculation:
let the two numbers be x and 6x
x × 6x = 29 × 174
x2 = 29 × 29
x = 29
Largest number is 6x,
= 6 × 29
= 174
Answer is 174.
17. In ∆ ABC, right-angled at B, AB = 24 cm, BC = 7 cm. The value of tan C is:
(a) 12/7
(b) 24/7
(c) 20/7
(d) 7/24
Answer: (b) 24/7
Explanation: AB = 24 cm and BC = 7 cm
tan C = Opposite side/Adjacent side
tan C = 24/7
18. The value of tan 60°/cot 30° is equal to:
(a) 0
(b) 1
(c) 2
(d) 3
Answer: (b) 1
Explanation: tan 60° = √3 and cot 30° = √3
Hence, tan 60°/cot 30° = √3/√3 = 1
19. sin (90° – A) and cos A are:
(a) Different
(b) Same
(c) Not related
(d) None of the above
Answer: (b) Same
Explanation: By trigonometry identities.
Sin (90°-A) = cos A {since 90°-A comes in the first quadrant of unit circle}
20. The value of sin 60° cos 30° + sin 30° cos 60° is:
(a) 0
(b) 1
(c) 2
(d) 4
Answer: (b) 1
Explanation: sin 60° = √3/2, sin 30° = ½, cos 60° = ½ and cos 30° = √3/2
Therefore,
(√3/2) x (√3/2) + (½) x (½)
= (3/4) + (1/4)
= 4/4
= 1
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