Q1. The reciprocal of (-3)/8×((-7)/13) is
(a) 104/21
(b) (-104)/21
(c) 21/104
(d) (-21)/104
Q3. Which of the numbers –20, -3/4,1/2, 10 is greater than its square?
(a) 1/2
(b) -3/4
(c) 10
(d) –20
Q4. The smallest value of ‘y’ in the number 71y8978 so that it is divisible by 11 is
(a) 3
(b) 1
(c) 0
(d) 2
Q6. The LCM of 121, 125 and 135 is
(a) 3³ × 5³ × 11²
(b) 3³ × 5² × 11²
(c) 3² × 5³ × 11³
(d) 3² × 5² × 11²
Q7. The additive inverse of S, where S = 1 – 2 + 3 – 4 + 5 – 6 + 7 – 8 + ……………… + 49 – 50, is:
(a) 1
(b) 0
(c) 25
(d) –25
Solutions
S1. Ans.(a)
Sol.
(-3)/8×(-7)/13=21/104
Reciprocal of (-3)/8×(-7)/13
= Reciprocal of 21/104
=104/21
S4. Ans.(a)
Sol. A number is divisible by 11 if the difference of sum of digits at even places and odd places is either 0 or divisible by 11.
Here, the difference of sum of digits at even places and odd places in 71y8978
= (7 + y + 9 + 8) – (1 + 8 + 7)
= 24 + y – 16
= y + 8
The smallest value of y for which y + 8 is divisible by 11 is 3.
S6. Ans.(a)
Sol. We have
121 = 11 × 11 = 11²
125 = 5 × 5 × 5 = 5³
135 = 3 × 3 × 3 × 5 = 3³ × 5
Therefore, the LCM of 121, 125 and 135 = 3³ × 5³ × 11²
S7. Ans.(c)
Sol. We have
S = 1 – 2 + 3 – 4 + 5 – 6 + 7 – 8 + ………. + 49 – 50
= (1 – 2) + (3 – 4) + (5 – 6) + (7 – 8) + ………. + (49 – 50);
Here, there would be 25 parentheses.
= (1 – 2) + (3 – 4) + (5 – 6) + (7 – 8) + ………. + (49 – 50);
= (–1) + (–1) + (–1) + (–1) + ……… + (–1);
Here, there would be 25 parentheses
= –25
Therefore, the additive inverse of S would be 25.
