Algebra Play Class 8 Solutions

Let the starting number be (x). The steps give x\to2x\to2x+4\to x+2\to2, so the starting number cancels. To end at 3, add 6 before dividing by 2; to end at 5, add 10. In general, doubling, adding 2k, halving, and subtracting the original number always gives k.

Example: choose (x), multiply by 3, add 12, divide by 3, add 7, then subtract (x). Algebraically this is (3x+12)/3+7-x=x+4+7-x=11. The trick always returns 11.

If the month is M and day is D, the displayed operations produce 100M+165+D. Subtract 165 from the reported answer to get 100M+D; the last two digits give the day and the preceding digits give the month.

Subtract 165 from each answer. (i) 1269 - 165 = 1104, so the date is 4 November. (ii) 394 - 165 = 229, so it is 29 February (valid in a leap year). (iii) 296 - 165 = 131, so it is 31 January.

Yes. The operations must end in 100M+C+D, where C is known. For example: multiply the month by 2, add 3, multiply by 50, then add the day. The result is 100M+150+D; subtract 150 to read M and D.

Each cell is the sum of the two below it. The first top is 8. The second has middle row 7,7 and top 14. The third has rows 9,9,5; then 18,14; and top 32.

The middle-right cell is 10 - 4 = 6. The bottom-middle cell is 4 - 1 = 3. The bottom-right cell is 6 - 3 = 3. Thus the rows are 1,3,3; then 4,6; then 10.

Let the bottom-middle value be c. The top is (12+c)+(c+8)=20+2c. Thus 20+2c=60, so c=20. The middle row is 32,28 and the top is 60.

Left pyramid: bottom 4,9,6,1; next 13,15,7; next 28,22; top 50. Middle: bottom 5,14,7,2; next 19,21,9; next 40,30; top 70. Right: bottom 3,5,5,2; next 8,10,7; next 18,17; top 35.

For two bottom values a,b, the top is a+b. For three values a,b,c, it is a+2b+c. The coefficients are Pascal-triangle coefficients because each interior value contributes along every upward path.

Use a+2b+c. The tops are 4+26+8=38, 7+22+3=32, and 10+28+25=63.

For bottom row a,b,c,d, the successive rows lead to top a+3b+3c+d. The coefficients 1,3,3,1 count the upward paths from each bottom cell.

Use a+3b+3c+d. For 8,19,21,13 the top is 141. For 7,18,19,6 it is 124. For 9,7,5,11 it is 56.

The bottom is 1,2,3; the next row is 3,5; and the top is 8. Every displayed number is in the sequence 1,2,3,5,8,... .

Adjacent sums are again sequence terms: F_i+F_{i+1}=F_{i+2}. For four rows the pyramid ends at F_7=21. For 29 rows it ends at F_{57}=591286729879, using the book's indexing F_1=1,F_2=2. Every entry remains a Virahanka-Fibonacci number.

If the bottom is F_1,F_2,\ldots,F_n, the next row is F_3,F_4,\ldots,F_{n+1}. Each upward step advances the starting index by 2. Thus all entries are sequence terms and the top is F_{2n-1}.

If the top-left date is a, the block is a,a+1,a+7,a+8, with sum 4a+16. Hence a=(S-16)/4. For sum 36, a=5, so the block is 5,6/12,13. A reported sum is possible only when these four dates fit the calendar.

Use three consecutive dates in one row: a,a+1,a+2. Their sum is 3a+3, so from a reported sum S, calculate a=(S-3)/3. For example, sum 42 gives a=13, hence 13,14,15.

Let the square be s and circle be c. Then 2s+c=27 and s+2c=21. Subtracting gives s-c=6; solving gives s=11,c=5. The last row c+s+c totals 5+11+5=21.

Let circle be c and diamond be d. The rows give c+2d=18 and d+2c=15, so d=7,c=4. The third row totals 7+4+4=15. The three column totals are 18,15,15.

The maximum is 32\times5=160. For digits p<q<r, only rq\times p, rp\times q, and qp\times r need comparison. Expanding the last two shows qp\times r=10qr+pr>10qr+pq=rp\times q, so the largest digit is the multiplier and the other two appear in descending order.

Apply the rule: largest digit as multiplier, remaining digits descending. Thus 31\times7=217 and 53\times9=477.

For digits a,b, the absolute difference is |10a+b-(10b+a)|=9|a-b|. It is therefore divisible by 9, and the quotient is the absolute difference between the two digits. If a>b, the unsigned difference is 9(a-b).

(10a+b)+(10b+a)=11a+11b=11(a+b). Therefore the sum is always divisible by 11, including examples 31+13=44 and 28+82=110.

abc+bca+cab=(100a+10b+c)+(100b+10c+a)+(100c+10a+b)=111(a+b+c)=3\times37(a+b+c). Hence it is always divisible by both 37 and 3.

abcabc=1000(abc)+abc=1001(abc), and 1001=7\times11\times13. Dividing by those three factors therefore leaves the original three-digit number (abc).

Let (x) be the starting flowers and k the equal offering. The remaining amounts after the first two shrines are 2x-k and 4x-3k; after the third doubling, all 8x-6k flowers equal k. Thus 8x=7k. The smallest positive whole-number solution is x=7,k=8: start with 7 and place 8 at each shrine.

If all 55 were hens, there would be 110 legs. The extra 40 legs come from horses replacing hens, adding 2 legs each, so there are 20 horses and 35 hens. Algebraically, h+r=55 and 2h+4r=150 gives the same result.

Let the daughter be d; the mother is 5d. Then 5d+6=3(d+6), so 2d=12 and d=6. The daughter is 6 and the mother is 30.

Let Gauri have g; Naina has 2g. After Naina gives 3 cows, 2g-3=g+3, so g=6. Gauri has 6 cows and Naina has 12.

Daily cost for 100 dosas is 5000+10(100)=6000. A ₹2000 profit needs ₹8000 revenue, so price = ₹80 per dosa. At a ₹50 price, profit on n dosas is 50n-(5000+10n)=40n-5000. Setting this to 2000 gives n=175 dosas.

Every fraction equals 1/3: 1/3, 4/12, and 9/27. In the nth term, the numerator is the sum of the first n odd numbers, n^2. The denominator is the sum of the next n odd numbers beginning at 2n+1, which is (2n)^2-n^2=3n^2. Their ratio is 1/3.

If the fee is 8, work backward from the 8 coins owed after the third doubling: before that doubling Karim had 4; before the second fee he had 12, so before doubling he had 6; before the first fee he had 14, so initially he had 7 coins. For initial amount (x) and fee c, the balance after three paid rounds is 8x-7c.

It exceeds (x) exactly when c<x. To take all the coins after three rounds, the genie sets c=8x/7, when this is a whole number.