Wipro Turbo Quiz on Progression

Unit digit in 3^4 = 1
Unit digit in (3^4)^17 = 1

Unit digit in 3^(69)
= Unit digit in [ (3^4)^17 x 3 ] = (1 x 3) = 3

Unit digit in 7^4
= Unit digit in (7^4)^17 is 1.

Unit digit in 7^(71)
= Unit digit in [(7^4)^17 x 7^3] = (1 x 3) = 3

Unit digit in 9^5 = 9
Required digit = Unit digit in (3 x 3 x 8) = 9

1^st Method:
8th term = a+7d = 39 ........... (i)
12th term = a+11d = 59 ........... (ii)
(i)-(ii);

Or, a+7d-a-11d = 39-59; Or, 4d = 20;
Or, d = 5;
Hence, a+7*5 = 39;
Thus, a = 39-35 = 4.
2nd Method (Thought Process):
8th term = 39;
And, 12th term = 59;
Here, we see that 20 is added to 8th term 39 to get 12th term 59 i.e. 4 times the common difference is added to 39.
So, CD = 20/4 = 5.
Hence, 7 times CD is added to 1st term to get 39. That means 4 is the 1st term of the AP.

Let, first term = a and normal contrast =d.
T6 = 12 ⇒ a+5d= 12 …. (i)
T8= 22 ⇒ a+7d = 22 … (ii)
On subtracting (i) from (ii), we get 2d = 10 ⇒ d = 5

Putting d= 5 in (i), we get a+5*5 = 12 ⇒ a= (12-25) =-13
∴ First term = - 13, normal distinction = 5.

T16= a+ 15d = - 13+15*5 = (75-13) = 62

Let the numbers be (a-d), an and (a+d). At that point,

(a-d) +a+ (a+d) = 15 ⇒ 3a = 15 ⇒ a = 5
(a-d)*a*(a+d) = 80 ⇒ (5-d)*5 * (5+d) = 80
⇒ (25-d2) = 16 = d2 =9 ⇒ d = 3
Numbers are 2, 5, 8 or 8, 5, 2.

Let us assume it take n time units to reach 512000 microbes, where each time unit = 2 minutes

Number of microbes in nth time unit=1000∗2^n

1000∗2^n=512000

n=9

Thus, total time taken = 9*2 = 18 minutes

The sum of an arithmetic series is the average of the first and last terms, times the number of terms.

For this series, the average of the first and last terms is (97+75)/2=86(97+75)/2=86

For any arithmetic series with a common difference of 1, the number of terms is the largest term minus the smallest term plus 1.

For this series, that is 97−75+1=2397−75+1=23

Therefore, the sum of this arithmetic series is 86⋅23=1978

Let, it be the n-th term of this GP series.

a = 5; r = 2.

1280 = 5*{2^(n - 1)}

{2^(n - 1)} = 256 = (2^8)

So, (n- 1) = 8; therefore, n = 9.

Let the required value be Rs a.
Also d= 500, n = 10 and Sn = 50000.
Sn = n/2 [2a + (n -1) d] ⇒ 10/2 *[2a + 9 *500] = 50000
⇒ 2a + 4500 = 10000⇒ 2a=5500⇒ a = 2750

LCM of 6 and 9: 18

>6=2*3

>9=3*3

What is the highest multiple of 18 under 100?

>200-100=100

90 = 18*5

Calculate the first starting position, 108

>108=18*6

If 108 + 90 < 200 AND 200–(108+90) < 18

then 90/18 is the answer.

The answer is 5

 

Ans: 24.

0 ,2, 9, ?, 50, 90, 147.

0,2,9,24,50,90,147 -->
diff b/w two terms
2-0=2,
9-2=7,
24-9=15,
50-24=26,
90-50=40,
147-90=57,

The new series : 2,7, 15, 26, 40, 57
diff & new series: 5, 8, 11, 14, 17
diff : 3, 3, 3, 3