Question
111
Mathematics
Find the value of k if the number 5487k2 is completely divisible by 11.
(A)
2
(B)
4
(C)
6
(D)
Correct Answer:
Explanation:
For divisibility by 11, the difference between the sums of digits in alternating positions must be a multiple of 11. For 5487k2, the difference is (5 + 8 + k) - (4 + 7 + 2) = k. Therefore, k must be 0 or a multiple of 11. Among the given options, k = 0 is correct.
For divisibility by 11, the difference between the sums of digits in alternating positions must be a multiple of 11. For 5487k2, the difference is (5 + 8 + k) - (4 + 7 + 2) = k. Therefore, k must be 0 or a multiple of 11. Among the given options, k = 0 is correct.
Question
112
Mathematics
The average age of 35 students in a class is 16 years. The average age of 21 students is 14 years. What is the average age of the remaining 14 students?
(A)
15 years
(B)
17 years
(C)
18 years
(D)
19 years
Correct Answer:
19 years
Explanation:
The total age of 35 students is 35 × 16 = 560 years. The total age of 21 students is 21 × 14 = 294 years. Therefore, the total age of the remaining 14 students is 560 - 294 = 266 years. Their average age is 266 ÷ 14 = 19 years.
The total age of 35 students is 35 × 16 = 560 years. The total age of 21 students is 21 × 14 = 294 years. Therefore, the total age of the remaining 14 students is 560 - 294 = 266 years. Their average age is 266 ÷ 14 = 19 years.
Question
113
Mathematics
Find the average of the first 100 natural numbers.
(A)
42
(B)
50.5
(C)
55
(D)
52
Correct Answer:
50.5
Explanation:
The first 100 natural numbers are 1 to 100. The average of an arithmetic sequence is (first term + last term) ÷ 2 = (1 + 100) ÷ 2 = 50.5.
The first 100 natural numbers are 1 to 100. The average of an arithmetic sequence is (first term + last term) ÷ 2 = (1 + 100) ÷ 2 = 50.5.
Question
114
Mathematics
600 gm of sugar solution has 40% sugar in it by weight. How much sugar must be added to make it a 50% sugar solution by weight?
(A)
100 gm
(B)
150 gm
(C)
120 gm
(D)
180 gm
Correct Answer:
120 gm
Explanation:
Sugar in 600 g solution = 40% × 600 = 240 g. If x grams of sugar are added, (240 + x)/(600 + x) = 50/100. Solving gives x = 120 g.
Sugar in 600 g solution = 40% × 600 = 240 g. If x grams of sugar are added, (240 + x)/(600 + x) = 50/100. Solving gives x = 120 g.
Question
115
Mathematics
The area of a trapezium is 336 sq. cm. If its parallel sides are in the ratio 5:7 and the perpendicular distance between them is 14 cm, then the smaller of the parallel sides is:
(A)
20 cm
(B)
25 cm
(C)
27.5 cm
(D)
28 cm
Correct Answer:
20 cm
Explanation:
Let the parallel sides be 5x and 7x. Area of the trapezium = 1/2 × (5x + 7x) × 14 = 336. Thus x = 4, so the smaller parallel side = 5 × 4 = 20 cm.
Let the parallel sides be 5x and 7x. Area of the trapezium = 1/2 × (5x + 7x) × 14 = 336. Thus x = 4, so the smaller parallel side = 5 × 4 = 20 cm.
Question
116
Mathematics
Two years ago, the ratio of the ages of Rohan and Ajit was 4:3. Two years from the present time, the ratio of their ages will become 5:4. Find the average of their present ages (in years).
(A)
32
(B)
14
(C)
15
(D)
16
Correct Answer:
16
Explanation:
Let the ages of Rohan and Ajit two years ago be 4x and 3x. Two years from now, their ages will be 4x+4 and 3x+4. Given (4x+4)/(3x+4) = 5/4, solving gives x = 4. Their present ages are 18 and 14 years. Average = (18+14)/2 = 16 years.
Let the ages of Rohan and Ajit two years ago be 4x and 3x. Two years from now, their ages will be 4x+4 and 3x+4. Given (4x+4)/(3x+4) = 5/4, solving gives x = 4. Their present ages are 18 and 14 years. Average = (18+14)/2 = 16 years.
Question
117
Mathematics
Ramya lent a certain sum of money at 4% simple interest to Seema and in 8 years the interest amounted to 3400 less than the sum lent. Find the sum lent by Ramya to Seema.
(A)
₹3050
(B)
₹6100
(C)
₹8200
(D)
₹5000
Correct Answer:
₹5000
Explanation:
Simple Interest = P × R × T / 100 = P × 4 × 8 / 100 = 0.32P. According to the question, the interest is ₹3400 less than the principal, so 0.32P = P - 3400. Thus 0.68P = 3400 and P = ₹5000.
Simple Interest = P × R × T / 100 = P × 4 × 8 / 100 = 0.32P. According to the question, the interest is ₹3400 less than the principal, so 0.32P = P - 3400. Thus 0.68P = 3400 and P = ₹5000.
Question
118
Mathematics
(A)
Only II
(B)
Only I
(C)
Both I and II
(D)
Neither I nor II
Correct Answer:
Only I
Explanation:
The product of two negative numbers is always positive. Therefore, (-16) × (-23) is positive. Similarly, (-19) × (-37) is also positive, not negative. Hence, only statement I is true.
The product of two negative numbers is always positive. Therefore, (-16) × (-23) is positive. Similarly, (-19) × (-37) is also positive, not negative. Hence, only statement I is true.
Question
119
Mathematics
A shopkeeper sells sugar in such a way that the selling price of 950 gm is the same as the cost price of one kilogram. Find his gain percent.
(A)
50/9%
(B)
50/19%
(C)
100/19%
(D)
100/21%
Correct Answer:
100/19%
Explanation:
Let the cost price of 1 kg of sugar be ₹100. Then the selling price of 950 g is ₹100. Therefore, the selling price of 1 kg is 1000 × 100/950 = 2000/19. Gain = 2000/19 - 100 = 100/19. Hence, gain percentage = 100/19%.
Let the cost price of 1 kg of sugar be ₹100. Then the selling price of 950 g is ₹100. Therefore, the selling price of 1 kg is 1000 × 100/950 = 2000/19. Gain = 2000/19 - 100 = 100/19. Hence, gain percentage = 100/19%.
Question
120
Mathematics
Enter the exact English question from the JPEG here.
(A)
1/5
(B)
1/10
(C)
2/5
(D)
1/2
Correct Answer:
1/5
Explanation:
Solution according to the question should be added here.
Solution according to the question should be added here.
Total Questions:
120
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