SSC CGL Tier1 Quantitative Aptitude Number System Test 1

Let S_n =1 + 2 + 3 + ... + 45. This is an A.P. in which a =1, d =1, n = 45.

S_n =$ \dfrac{n}{2} $[2$ a $ + $\left( n - 1\right)$ d ]=$ \dfrac{45}{2} $ x [2 x 1 + (45 - 1) x 1]=$ \dfrac{45}{2} $x 46= 45 x 23

= 45 x (20 + 3)

= 45 x 20 + 45 x 3

= 900 + 135

= 1035.

Shorcut Method:

S_n =$ \dfrac{n(n + 1)}{2} $=$ \dfrac{45(45 + 1)}{2} $= 1035.

(11² + 12² + 13² + ... + 20²) = (1² + 2² + 3² + ... + 20²) - (1² + 2² + 3² + ... + 10²)

Ref: (1² + 2² + 3² + ... + n²) =$\dfrac{1}{6} n(n + 1)(2n + 1)$    

= (2870 - 385)

= 2485.

Required sum = $\left(2 + 4 + 6 + ... + 30\right)$

This is an A.P. in which $ a $ = 2, $ d $ = $\left(4 - 2\right)$ = 2 and $ l $ = 30.

Let the number of terms be $ n $. Then,

t_n = 30 $\Rightarrow$    $ a $ + $\left( n - 1\right)$$ d $ = 30

$\Rightarrow$ 2 + $\left( n - 1\right)$ x 2 = 30

$\Rightarrow n $ - 1 = 14

$\Rightarrow n $ = 15

$\therefore$S_n =$ \dfrac{n}{2} $$\left(a + l\right )$=$ \dfrac{15}{2} $ x $\left(2 + 30\right)$   = 240.

[Place value of 6] - [Face value of 6] = 6000 - 6 = 5994

The sum of two odd number is even. So, a + b is even.

This is an A.P. in which $ a $ = 6, $ d $ = 6 and S_n = 1800

Then,$ \dfrac{n}{2} $[2$ a $ +$\left ( n - 1\right)$$ d $] = 1800

$\Rightarrow$ $ \dfrac{n}{2} $[2 x 6 + $\left( n - 1\right)$ x 6] = 1800

$\Rightarrow$ 3$ n $ $\left(n + 1\right)$ = 1800

$\Rightarrow n \left(n + 1\right)$ = 600

$\Rightarrow n $² + $ n $ - 600 = 0

$\Rightarrow n $² + 25$ n $ - 24$ n $ - 600 = 0

$\Rightarrow n $$\left( n + 25\right)$ - 24$\left( n + 25\right)$ = 0

$\Rightarrow$ $\left( n + 25\right)$$\left( n - 24\right)$ = 0

$\Rightarrow n $ = 24

Number of terms = 24.

Unit digit in 7⁹⁵ = Unit digit in [(7⁴)²³ x 7³]

= Unit digit in [Unit digit in 2401 ²³] x (343)]

= Unit digit in (1²³ x 343)
= Unit digit in (343)
= 3

Unit digit in 3⁵⁸ = Unit digit in [(3⁴)¹⁴ x 3²]
= Unit digit in [Unit digit in (81)¹⁴ x 3²]
= Unit digit in [(1)¹⁴ x 3²]
= Unit digit in 1 x 9
= Unit digit in 9
= 9

Unit digit in (7⁹⁵ - 3⁵⁸) = Unit digit in (343 - 9) = Unit digit in (334) = 4.

So, Option b is the answer.

Sum of the cubes of first n natural numbers = $\left[\dfrac{n × (n + 1)}{2}\right]^{2}$
Sum of the cubes of first 12 numbers = $\left[\dfrac{12 × (12 + 1)}{2}\right]^{2}$
= $\left[\dfrac{12 × 13}{2}\right]^{2}$
= $(6 × 13)^{2}$
= $78^{2}$
= 6084

Sum of squares of n numbers is given by the formula $\dfrac{n(n+1)(2n+1)}{6}$
Here n = 20 so,
= $\dfrac{35 × (35 + 1)(2×35 + 1)}{6}$
= $\dfrac{35 × 36 × 71}{6}$
= 210 × 71
= 14910
So the sum of squares of first 35 terms is 14910.

Sum of the squares of first n odd numbers = $\dfrac{n(2n + 1)(2n - 1)}{3}$
Sum of squares of first 10 odd numbers = $\dfrac{10 × (2 × 10 + 1)(2 × 10 - 1)}{3}$
= $\dfrac{10 × 21 × 19}{3}$
= 70 × 19
= 1330