SSC CGL 20201)What is the smallest interger that is a multiple of 5, 8 and 15 ?
120
LCM of 5, 8, 15 is 120
SSC CGL 20192)Let a, b and c be the fractions such that a < b < c. If c is divided by a, the result is \({5\over2}\), which exceeds b by \({7\over4}\).
If a + b + c = 1\({11 \over 12}\), then (c - a) will be equal to :
\({1 \over 2}\)
ATQ,
\(\frac{c}{a} = \frac{5}{2};\)
c = \(\frac{5a}{2};\)
b = \(\frac{5}{2} - \frac{7}{4} = \frac{3}{4};\)
a + b + c = \(1\frac{11}{12} = \frac{23}{12};\)
a +\( \frac{3}{4} + \frac{5a}{2} = \frac{23}{12};\)
\(\frac{7a}{2} = \frac{23}{12} - \frac{3}{4};\)
7a = \( \frac{23}{6} - \frac{3}{2};\)
7a = \( \frac{7}{3} \);
a = \( \frac{1}{3} \);
c =\( \frac{5}{2} \times \frac{1}{3} = \frac{5}{6};\)
c - a =\( \frac{5}{6} - \frac{1}{3} = \frac{1}{2}\)
SSC CGL 20193)If a nine-digit number 389x 6378y is divisible by 72, then the value of \(\sqrt{6x+7y}\) will be :
8
389x6378y is divisible by 72,
Factor of 72 = \(8 \times 9;\)
So, number is divisible by 8 and 9 both.
Divisibility rule for 8,
78y (last three digits should be divisible by 8);
784 is divisible by 8 so,
Value of y = 4;
Divisibility rule of 9,
3 + 8 + 9 + x + 6 + 3 + 7+ 8 + 4
= 48 + x;
54 is divisible by 9;
So, x = 54 - 48 = 6;
Value of\( \sqrt{6x + 7y} = \sqrt{6 \times 6 + 7 \times 4} = \sqrt{36 + 28} =\sqrt{64} = 8\)
SSC CGL 20194)When 12, 16, 18, 20 and 25 divide the least number x, the remainder in each case is 4 but x is divisible by 7. What is the digit at the thousand's place in x ?
8
Number = (LCM of 12, 16, 18, 20 and 25)k + 4
= 3600k + 4;
The number should be divisible by the 7 so,
Value of K = 5;
Number = \(3600 \times 5 + 4 \)= 18000 + 4 = 18004;
The digit at the thousands’ place = 8
SSC CGL 20195)When 7897, 8110 and 8536 are divided by the greatest number x, then the remainder in each case is the same. The sum of the digits of x is:
6
Let the remainder be k. 7897 - k = ax ; 8110 - k = bx; 8536 - k = cx ; Common factor is x. So difference between the numbers, 8110 - 7897 = 213 ; 8536 - 8110 = 426; 8536 - 7897 = 639 ; HCF of 213, 426 and 639 is 213. x = 213; Sum of the digits of x = 2 + 1 + 3 = 6
SSC CGL 20196)One of the factors of\( (8^{2k} + 5^{2k})\), where k is an odd number, is:
89
\((8^{2k} + 5^{2k})\),k is odd nuber so,
Let the k be 1.
=\((8^{2} + 5^{2})\)
= 64 + 25 = 89
SSC CGL 20197)Let \(x= (633)^{24} - \)\((277)^{38} + (266)^{54}\). What is the units digit of x ?
8
x = \((633)^{24} - (277)^{38} + (266)^{54}\)For the unit digit,
24 = 4 \times 6 + 0(remainder);
38 = 4 \times 9 + 2(remainder);
54 = 4 \times 13 + 2(remainder);
Now,
(Base number unit digit)^{remainder}
= \((3)^0 - (7)^2 + (6)^2\);
On consider unit digit,
= 1 - 9 + 6 = 7 - 9;
or 17 - 9 = 8;
8 is the units digit of x.
SSC CGL 20198)The sum of the digits of a two-digit number is \( \frac{1}{7} \) of the number. The units digit is 4 less than the tens digit. If the number obtained on reversing its digits is divided by 7, the remainder will be:
6
Let the number be (10a + b).
ATQ,
a + b =\( \frac{10a + b}{7}\);
7a + 7b = 10a + b;
6b = 3a;
2b = a ---(1);
a - b = 4 ---(2);
From eq (1) and (2),
2b - b = 4;
b = 4;
a = \(4 \times 2
\)= 8;
Number = 10a + b = \(10 \times 8 + 4 =\) 84;
reverse of the number = 48;
Remainder after divide by 7 = 48/7 = 6
SSC CGL 20199)If x is the remainder when \(3^{61284}\) is divided by 5 and y is the remainder when \(4^{96}\) is divided by 6, then what is the value of \((2x-y)\) ?
-2
x is the remainder when \(3^{61284}\) is divided by 5; So, \({3^{61284}\over5}={3^{4\times15321}\over5}\) ; \({3^4\over5}={81\over5}\); Remainder = 1; x=1;
y is the remainder when \(4^{96}\) is divided by 6; remainder is always '4'; thereforey = 4.
2x-y = 2-4 = -2
SSC CGL 201910)The LCM of two numbers x and y is 204 times their HCF. If their HCF is 12 and the difference between the numbers is 60, then x + y = ?
348