SNAP Progressions and Series Questions

Number of two-digit numbers divisible by 5 = 10, 15, 20,......95

Number of two-digit numbers divisible by 7 = 14, 21, 28,.....98

Number of two-digit numbers divisible by 35 = 35, 70

Using $$S_{n}=\ \dfrac{n}{2}\left(2a+\left(n-1\right)d\right)$$

Sum of the numbers divisible by 5 $$=\dfrac{18}{2}\left(10+95\right)=9\times105=945$$

Sum of the numbers divisible by 7 $$=\dfrac{13}{2}\left(14+98\right)=13\times56=728$$

We need to subtract the numbers divisible by both 5 and 7 $$=35+70=105$$. 

So total $$=945+728-105=1568$$

Natural numbers less than 201 and divisible by 5 are 5, 10, . . .200. Here, 5+200 = 10+195  . . = 100+105. Hence, there are 40/2 = 20 pairs totalling 205.

Sum of all natural numbers less than 201 divisible by 5 is 20*205 = 4100

Similarly, natural numbers less than 201 and divisible by 6 are 6, 12, . . .  198. 6+198= 12+192= . . . = 96+108  and 102. Hence, there are $$\lfloor201/6 /2 \rfloor$$ = 16 such pairs and 102.

Sum of all natural numbers less than 201, divisible by 6 is 16*204 + 102 = 3366

Similarly, Sum of all natural numbers divisible by both 5 and 6 is 6*210/2 = 630. These numbers have been summed in both the above totals. Hence, we need to deduct 2 times 630.

So, required total is 4100+3366-2*630 = 6206

For the 1st drop the distance travelled by the ball = 200m

After 1st drop it rebounces to a height of 4/5 * 200 and then falls from that height . The total distance travelled in this case would be $$2\cdot\frac{4}{5}\cdot200$$

Similarly the total distance travelled in next case would be $$2\cdot\frac{4}{5}\cdot\frac{4}{5}\cdot200$$

So total distance = 

200 + $$2\cdot200\cdot\left(\frac{4}{5}+\left(\frac{4}{5}\right)^2+\left(\frac{4}{5}\right)^{3\ }+....\right)$$

200+400$$\left(\frac{4}{\frac{5}{1-\frac{4}{5}}}\right)$$    = 1800m

The numbers will be of the form 7k+2, where k is a whole number.

The smallest two-digit number is when k=2, which is 16, and the largest 2 digit number is 93 when k = 13.

So sum = 16+23+.... +93, which is in AP. 

Sum to n terms of an AP= n/2(a+l), where n=number of terms , a=1st term ,l=last term

Here n=12 , a=16 , l=93

Hence sum =  $$\frac{12}{2}\left(16+93\right)$$  = 654

It is a decreasing AP 25,24.5,24,23.5,...

The sum will be largest, if nth term is zero then

25 + (n-1)*(-.5) = 0

=> n = 51

after 51 terms AP contains -ve terms

so, max sum will be obtained upto 50/51th term

S = $$\frac{51}{2}$$ *(25+0) = 637.5

The first natural number between 10 and 300 which is divisible by 9 is 18

The last natural number which is divisible by 9 and is less than 300 is 297

This is an AP with first term 18, the last term 297 and common difference = 9

297=18+(n-1)9

279=(n-1)9

n-1 = 31

n=32

A is the correct answer.

The price of Darjeeling Tea (in rupees per kilogram) is 100 + 0.1n till 100th day and remains constant

The price of Ooty tea (in rupees per kilogram) is 85 + 0.15n

Price of Darjeeling tea on 100th day = 100 + 0.1(100) = 110

Price of Ooty tea on 100th day = 85 + 0.15(100) = 100

Their prices are not equal

Therefore, 110 = 85 + 0.15n

                      n = $$\frac{500}{3}$$ = 166.67

Price will be equal on 167th day

January(31) + February(28) + March(31) + April(30) + May(31) = 151 days

June 16th is 167th day.

Answer is option B

Let the first term of the geometric progression be a and common ratio be r 
Now as per given condition :
$$a+ar= a(1+r) =12$$  (1)
$$ar^2+ar^3 = 48$$ (2)
we get $$ar^2(1+r) =48$$  (3)
Dividing (1) and (3)
we get :
$$1/r^2 = 1/4$$
we get $$1/r = 1/2  or -1/2 $$
but since terms are alternatively positive and negative
we get $$r = -2$$
Substituting r value in Eq. 1,
we get $$a(-1) =12 $$
we get $$a=-12$$

Let n pages be missing starting from the p th page.

So,$$p+(p+1)+(p+2)\dots..+[p+(n−2)]+[p+(n−1)]=9808$$

$$9808=\frac{n}{2}[2\times p+(n−1)\times1]$$
$$n^2+(2p-1)n−19616=0$$
Now both p and n are natural numbers 
Now possible cases 
$$n=19616,\ p=−9807$$
$$n=−19616,\ p=9808$$
$$n=613,\ p=−29$$
$$n=−613,\ p=29$$
$$n=1,\ p=9808$$
$$n=−1,\ p=−9807$$
$$n=32,\ p=291$$
$$n=−32,\ p=−290$$

But since p and n has to be natural numbers, so all solutions except 5 and 7 are discarded.

Also, it is mentioned in the question that "a number of consecutive pages are missing". Thus, case 5 can be discarded(i.e. n = 1 and p = 9808)

So, the correct option is C

Here, $$\ \frac{\ n\left(n-1\right)}{2}$$(not inclusive) term to $$\ \frac{\ n\left(n+1\right)}{2}$$(inclusive) term is the nth place alphabet.   (1st place alphabet is a, 2nd place alphabet is b, and so on.) 

=> $$\ \frac{\ n\left(n-1\right)}{2}$$ < 288 < $$\ \frac{\ n\left(n+1\right)}{2}$$

=> n(n-1) < 576 < n(n+1)

n = 24 

At 24th place, x is the alphabet.

$$9^{th}$$ term = D

$$16^{th}$$ term = J

$$24^{th}$$ term = Y

$$27^{th}$$ term = S

No meaningful word can be made from the terms hence X is the answer

The identical terms will be of the form k*LCM(d1,d2) +smallest common number .

Where d1 , d2 are common differences in two series and smallest common term in two series = 6

So N= k*LCM ( 2,3) + 6

N= 6k+6   (where k is a whole number)

Since the last number in 1st series is 200 and last number in 2nd series is 300, it is sufficient to check multiples of till 200

Therefore total number of multiples of 6 below 200 = 33

Alternate Explanation:

S1 = 2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + .......198 + 200 (100 terms)

S2 = 3 + 6 + 9 + 12 + 18 + 21 + 24 + 27 .......+ 297 + 300 (100 terms)

Common terms forms an AP, as both series are also in AP

Common terms are : 6, 12, 18, 24, .......... 198 (As S1 is upto 200)

Now, first term a= 6

Common difference = 6

Last term = 198

Number of terms = ?

Last term = first term + (n-1)*common difference

198 = 6 + (n-1)*6

n = 33

For every cycle of 2 days his overall earnings will be 5 Rs ,

In 8 cycles i.e in 16 days he will earn 40 Rs, on 17th day he will earn 20 more rupees and his overall savings will reach to 60 Rs.